S K Mondal’s
Strength of Materials Contents Chapter – 1: Stress and Strain Chapter  2 : Principal Stress and Strain Chapter  3 : Moment of Inertia and Centroid Chapter  4 : Bending Moment and Shear Force Diagram Chapter  5 : Deflection of Beam Chapter  6 : Bending Stress in Beam Chapter  7 : Shear Stress in Beam Chapter  8 : Fixed and Continuous Beam Chapter  9 : Torsion Chapter10 : Thin Cylinder Chapter11 : Thick Cylinder Chapter12 : Spring Chapter13 : Theories of Column Chapter14 : Strain Energy Method Chapter15 : Theories of Failure Chapter16 : Riveted and Welded Joint
Er. S K Mondal IES Officer (Railway), GATE topper, NTPC ET2003 batch, 12 years teaching experienced, Author of Hydro Power Familiarization (NTPC Ltd)
Page 1 of 429
Note “Asked Objective Questions” is the total collection of questions from:20 yrs IES (20101992) [Engineering Service Examination] 21 yrs. GATE (20111992) and 14 yrs. IAS (Prelim.) [Civil Service Preliminary]
Copyright © 2007 S K Mondal
Every effort has been made to see that there are no errors (typographical or otherwise) in the material presented. However, it is still possible that there are a few errors (serious or otherwise). I would be thankful to the readers if they are brought to my attention at the following email address: swapan_mondal_01@yahoo.co.in S K Mondal
Page 2 of 429
1.
Stress and Strain
Theory at a Glance (for IES, GATE, PSU) 1.1 Stress (ı) When a material is subjected to an external force, a resisting force is set up within the component. The internal resistance force per unit area acting on a material or intensity of the forces distributed over a given section is called the stress at a point.
x
It uses original cross section area of the specimen and also known as engineering stress or conventional stress. Therefore, T
x
P A
P is expressed in Newton (N) and A, original area, in square meters (m2), the stress ǔ will be expresses in N/ m2. This unit is called Pascal (Pa).
x
As Pascal is a small quantity, in practice, multiples of this unit is used. 1 kPa = 103 Pa = 103 N/ m2 1 MPa =
106
Pa =
106 N/
m2
(kPa = Kilo Pascal) =1
N/mm2
1 GPa = 109 Pa = 109 N/ m2
(MPa = Mega Pascal) (GPa = Giga Pascal)
Let us take an example: A rod 10 mm q 10 mm crosssection is carrying an axial tensile load 10 kN. In this rod the tensile stress developed is given by
Tt
x
P 10 kN 10q103 N 100N/mm2 100MPa 2 A 10 mm q10 mm 100 mm The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis.
x
The force intensity on the shown section is defined as the normal stress. %F P T lim and Tavg %Al 0 %A A
x
Tensile stress (ıt)
If ǔ > 0 the stress is tensile. i.e. The fibres of the component tend to elongate due to the external force. A member subjected to an external force tensile P and tensile stress distribution due to the force is shown in the given figure.
Page 3 of 429
Chapter1
x
Stress and Strain
S K Mondalâ€™s
Compressive stress (Äąc)
If Ç” < 0 the stress is compressive. i.e. The fibres of the component tend to shorten due to the external force. A member subjected to an external compressive force P and compressive stress distribution due to the force is shown in the given figure.
x
Shear stress ( U )
When forces are transmitted from one part of a body to other, the stresses developed in a plane parallel to the applied force are the shear stress. Shear stress acts parallel to plane of interest. Forces P is applied transversely to the member AB as shown. The corresponding internal forces act in the plane of section C and are called shearing forces. The corresponding average shear stress U
P Area
1.2 Strain (Ä°) The displacement per unit length (dimensionless) is known as strain.
x
Tensile strain ( F t)
The elongation per unit length as shown in the figure is known as tensile strain. Ç†t = ĆŚL/ Lo It is engineering strain or conventional strain. Here we divide the elongation to original length not actual length (Lo + % L) Let us take an example: A rod 100 mm in original length. When we apply an axial tensile load 10 kN the final length of the rod after application of the load is 100.1 mm. So in this rod tensile strain is developed and is given by
Ft
x
%L L Lo 100.1mm 100 mm 0.1mm 0.001 (Dimensionless)Tensile Lo Lo 100 mm 100 mm
Compressive strain ( F c) If the applied force is compressive then the reduction of length per unit length is known as compressive strain. It is negative. Then Ä°c = (ÇťL)/ Lo
Let us take an example: A rod 100 mm in original length. When we apply an axial compressive load 10 kN the final length of the rod after application of the load is 99 mm. So in this rod a compressive strain is developed and is given by Page 4 of 429
Chapter1
Fc
x
Stress and Strain
S K Mondalâ€™s
%L L Lo 99 mm 100 mm 1mm 0.01 (Dimensionless)compressive 100 mm 100 mm Lo Lo Shear Strain ( H ): When a force P is applied tangentially to the element shown. Its edge displaced to dotted line. Where
E is
the lateral displacement of
the upper face of the element relative to the lower face and L is the distance between these faces. Then the shear strain is
(H )
E L
Let us take an example: A block 100 mm Ă— 100 mm base and 10 mm height. When we apply a tangential force 10 kN to the upper edge it is displaced 1 mm relative to lower face. Then the direct shear stress in the element (U )
10 kN 10q103 N 1 N/mm2 1 MPa 100 mmq100 mm 100 mmq100 mm
And shear strain in the element ( H ) =
1mm 0.1 Dimensionless 10 mm
1.3 True stress and True Strain The true stress is defined as the ratio of the load to the cross section area at any instant.
x
TT
load Instantaneous area
Where
T
and
F
T 1 F
is the engineering stress and engineering strain respectively.
True strain L
FT Â¨ Lo
Â?LÂŹ Â?A ÂŹ Â?d ÂŹ dl ln ÂžÂžÂž ÂÂÂ ln 1 F ln ÂžÂž o ÂÂÂ 2ln ÂžÂž o ÂÂÂ ÂžÂ&#x; Lo ÂŽÂ Â&#x;Âž A ÂŽ Â&#x;Âž d ÂŽ l
or engineering strain ( F ) = eFT 1 The volume of the specimen is assumed to be constant during plastic deformation. [ ' Ao Lo AL ] It is valid till the neck formation.
x
Comparison of engineering and the true stressstrain curves shown below
Page 5 of 429
Chapter1
Stress and Strain
S K Mondal’s
x
The true stressstrain curve is also known as the flow curve.
x
True stressstrain curve gives a true indication of deformation characteristics because it is based on the instantaneous dimension of the specimen.
x
In engineering stressstrain curve, stress drops down after necking since it is based on the original area.
x
In true stressstrain curve, the stress however increases after necking since the crosssectional area of the specimen decreases rapidly after necking.
x
The flow curve of many metals in the region of uniform plastic deformation can be expressed by the simple power law. ǔT = K(ǆT)n
Where K is the strength coefficient n is the strain hardening exponent n = 0 perfectly plastic solid n = 1 elastic solid For most metals, 0.1< n < 0.5
x
Relation between the ultimate tensile strength and true stress at maximum load The ultimate tensile strength Tu
Pmax Ao
The true stress at maximum load Tu T
Pmax A
Ao ¬ A ®
And true strain at maximum load F T ln Eliminating Pmax we get , Tu T
or
Ao e FT A
Pmax Pmax Ao q Tu e FT A Ao A
Where Pmax = maximum force and Ao = Original cross section area A = Instantaneous cross section area Let us take two examples: (I.) Only elongation no neck formation In the tension test of a rod shown initially it was Ao = 50 mm2 and Lo = 100 mm. After the application of load it’s A = 40 mm2 and L = 125 mm. Determine the true strain using changes in both length and area. Answer: First of all we have to check that does the
(If no neck formation
member forms neck or not? For that check Ao Lo AL
occurs both area and
or not?
gauge length can be used
Here 50 × 100 = 40 × 125 so no neck formation is
for a strain calculation.)
there. Therefore true strain
Page 6 of 429
Chapter1
Stress and Strain L
S K Mondal’s
125 ¬ dl ln 0.223 100 ® l
FT ¨ Lo
Ao ¬ 50 ¬ ln 0.223 40 ® A®
FT ln
(II.) Elongation with neck formation A ductile material is tested such and necking occurs then the final gauge length is L=140 mm and the final minimum cross sectional area is A = 35 mm2. Though the rod shown initially it was Ao = 50 mm2 and Lo = 100 mm. Determine the true strain using changes in both length and area.
Answer: First of all we have to check that does the
(After necking, gauge
member forms neck or not? For that check Ao Lo AL
length gives error but
or not?
area and diameter can
Here AoLo = 50 × 100 = 5000 mm3 and AL=35 × 140
be
= 4200 mm3. So neck formation is there. Note here
calculation of true strain
AoLo > AL.
at fracture and before
Therefore true strain
fracture also.)
used
for
the
Ao ¬ 50 ¬ ln 0.357 35 ® A ®
FT ln L
But not FT ¨ Lo
140 ¬ dl ln 0.336 (it is wrong) 100 ® l
1.4 Hook’s law According to Hook’s law the stress is directly proportional to strain i.e. normal stress (ǔ) B normal strain (ǆ) and shearing stress ( U ) B shearing strain ( H ). ǔ = Eǆ and U GH The coefficient E is called the modulus of elasticity i.e. its resistance to elastic strain. The coefficient G is called the shear modulus of elasticity or modulus of rigidity.
1.5 Volumetric strain Fv
A relationship similar to that for length changes holds for threedimensional (volume) change. For volumetric strain, Fv , the relationship is Fv = (VV0)/V0 or Fv = ƦV/V0
P K
x
Where V is the final volume, V0 is the original volume, and ƦV is the volume change.
x
Volumetric strain is a ratio of values with the same units, so it also is a dimensionless quantity.
Page 7 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
x
ĆŚV/V= volumetric strain = Ç†x +Ç†y + Ç†z = Ç†1 +Ç†2 + Ç†3
x
Dilation: The hydrostatic component of the total stress contributes to deformation by changing the area (or volume, in three dimensions) of an object. Area or volume change is called dilation and is positive or negative, as the volume increases or decreases, respectively. e
p Where p is pressure. K
1.6 Youngâ€™s modulus or Modulus of elasticity (E) =
PL Äą = AÄŻ Â?
W PL 1.7 Modulus of rigidity or Shear modulus of elasticity (G) = = J AG
1.8 Bulk Modulus or Volume modulus of elasticity (K) =
'p 'v v
'p 'R R
1.10 Relationship between the elastic constants E, G, K, Îź
E
2G 1 P
9KG 3K G
3K 1 2P
[VIMP]
Where K = Bulk Modulus, N = Poissonâ€™s Ratio, E= Youngâ€™s modulus, G= Modulus of rigidity
x
For a linearly elastic, isotropic and homogeneous material, the number of elastic constants required to relate stress and strain is two. i.e. any two of the four must be known.
x
If the material is nonisotropic (i.e. anisotropic), then the elastic modulii will vary with additional stresses appearing since there is a coupling between shear stresses and normal stresses for an anisotropic material.
Let us take an example: The modulus of elasticity and rigidity of a material are 200 GPa and 80 GPa, respectively. Find all other elastic modulus.
2G 1 P 3K 1 2 P
Answer: Using the relation E
9KG we may find all other elastic modulus 3K G
easily Poissonâ€™s Ratio ( P ) : Bulk Modulus (K) :
1 P
3K
E 2G
E 1 2P
Â&#x;P
Â&#x;K
E 200 1 1 0.25 2G 2 u 80
E 3 1 2 P
200 133.33GPa 3 1 2 u 0.25
1.11 Poissonâ€™s Ratio (Îź) =
Transverse strain or lateral strain Â?y = Longitudinal strain Â?x
(Under unidirectional stress in xdirection)
x x
The theory of isotropic elasticity allows Poisson's ratios in the range from 1 to 1/2. Poisson's ratio in various materials
Page 8 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
Material
Poisson's ratio
Material
Poisson's ratio
Steel
0.25 â€“ 0.33
Rubber
0.48 â€“ 0.5
C.I
0.23 â€“ 0.27
Cork
Nearly zero
Concrete
0.2
Novel foam
negative
x
We use cork in a bottle as the cork easily inserted and removed, yet it also withstand the pressure from within the bottle. Cork with a Poisson's ratio of nearly zero, is ideal in this application.
1.12 For biaxial stretching of sheet Â§L Âˇ Â?1 ln Â¨ f 1 Â¸ ÂŠ Lo1 Âš Â§L Âˇ Â?2 ln Â¨ f 2 Â¸ ÂŠ Lo 2 Âš
Lo Original length L f Final length
Final thickness (tf) =
Initial thickness(t o ) eÂ?1 u eÂ?2
1.13 Elongation
x
A prismatic bar loaded in tension by an axial force P For a prismatic bar loaded in tension by an axial force P. The elongation of the bar can be determined as
PL AE
G
Let us take an example: A Mild Steel wire 5 mm in diameter and 1 m long. If the wire is subjected to an axial tensile load 10 kN find its extension of the rod. (E = 200 GPa)
PL AE
Answer: We know that G
Here given, Force (P) 10 kN Length(L) 1 m Area(A)
10 u 1000N
Sd2
S u 0.005
4
4
Modulous of Elasticity (E ) Therefore Elongation(G )
2
m2
1.963 u 105 m2
200 GPa 200 u 109 N/m2 PL AE
10 u 1000 u 1
1.963 u10 u 200 u 10
5
2.55 u 103 m
x
9
m
2.55 mm
Elongation of composite body
Elongation of a bar of varying cross section A1, A2,,An of lengths l1, l2,ln respectively.
G
l Âş P ÂŞ l1 l2 l3 n Âť ÂŤ E ÂŹ A1 A2 A3 An Âź Page 9 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
Let us take an example: A composite rod is 1000 mm long, its two ends are 40 mm2 and 30 mm2 in area and length are 300 mm and 200 mm respectively. The middle portion of the rod is 20 mm2 in area and 500 mm long. If the rod is subjected to an axial tensile load of 1000 N, find its total elongation. (E = 200 GPa). Answer: Consider the following figure
Given, Load (P) =1000 N Area; (A1) = 40 mm2, A2 = 20 mm2, A3 = 30 mm2 Length; (l1) = 300 mm, l2 = 500 mm, l3 = 200 mm E = 200 GPa = 200 u 109 N/m2 = 200 u 103 N/mm2 Therefore Total extension of the rod
G
P ÂŞ l1 l 2 l 3 Âş ÂŤ Âť E ÂŹ A1 A2 A3 Âź ÂŞ 300 mm 500 mm 200 mm Âş 1000 N uÂŤ Âť 3 2 200 u 10 N / mm ÂŹ 40 mm 2 20 mm 2 30 mm 2 Âź 0.196mm
x
Elongation of a tapered body
Elongation of a tapering rod of length â€˜Lâ€™ due to load â€˜Pâ€™ at the end
ÄŻ=
4PL S Ed1 d 2
(d1 and d2 are the diameters of smaller & larger ends)
You may remember this in this way, ÄŻ=
PL Â§S Âˇ E Â¨ d1 d 2 Â¸ ÂŠ4 Âš
i.e.
PL EA eq
Let us take an example: A round bar, of length L, tapers uniformly from small diameter d1 at one end to bigger diameter d2 at the other end. Show that the extension produced by a tensile axial load P is ÄŻ =
4PL . S d1 d 2 E
If d2 = 2d1, compare this extension with that of a uniform cylindrical bar having a diameter equal to the mean diameter of the tapered bar.
Answer: Consider the figure below d1 be the radius at the smaller end. Then at a X cross section XX located at a distance Ă— from the smaller end, the value of diameter â€˜dxâ€™ is equal to
Page 10 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
dx 2
d1 x Â§ d 2 d1 Â· Â¨ Â¸ 2 LÂ© 2 2 Â¹ x or d x d1 d 2 d1
L d1 1 kx
d 2 d1 1 u L d1
Where k
We now taking a small strip of diameter 'd x 'and length 'd x 'at section XX . Elongation of this section 'd x ' length d G
PL AE
P .dx Â§ S d x2 Â· Â¨ Â¸uE Â© 4 Â¹
4P .dx 2
S .^d1 1 kx ` E
Therefore total elongation of the taper bar x L
G
Â³ d G
4P dx
Â³ S Ed 1 kx
2 1
x 0
2
4PL S E d1d 2 Comparison: CaseI: Where d2 = 2d1 Elongation G I
4PL S Ed1 u 2d1
2PL S Ed12
Case â€“II: Where we use Mean diameter
dm
d1 d 2 2
d1 2d1 2
3 d1 2
Elongation of such bar G II
Extension of taper bar Extension of uniform bar
PL AE
2 16 9
Page 11 of 429
9 8
P .L
S Â§3 Â·
2
Â¨ d1 Â¸ .E 4Â©2 Â¹ 16PL 9S Ed12
Chapter1
x
Stress and Strain
S K Mondal’s
Elongation of a body due to its self weight (i) Elongation of a uniform rod of length ‘L’ due to its own weight ‘W’
WL į= 2AE The deformation of a bar under its own weight as compared to that when subjected to a direct axial load equal to its own weight will be half. (ii) Total extension produced in rod of length ‘L’ due to its own weight ‘ X ’ per with
į=
length.
Z L2 2EA
(iii) Elongation of a conical bar due to its self weight
į=
1.14
U gL2
WL 2 Amax E
6E
Structural members or machines must be designed such that the working stresses are less
than the ultimate strength of the material.
Working stress V w
Vy n
V ult n1
½ n=1.5 to 2 ° ° ¾ factor of safety n1 2 to 3 ° °¿
Vp n
1.15 Factor of Safety: (n) =
Vp
Proof stress
V y or V p or V ult Vw
1.16 Thermal or Temperature stress and strain
x
When a material undergoes a change in temperature, it either elongates or contracts depending upon whether temperature is increased or decreased of the material.
x
If the elongation or contraction is not restricted, i. e. free then the material does not experience any stress despite the fact that it undergoes a strain.
x
The strain due to temperature change is called thermal strain and is expressed as,
H x
D 'T
Where ǂ is coefficient of thermal expansion, a material property, and ƦT is the change in temperature.
x
The free expansion or contraction of materials, when restrained induces stress in the Page 12 of 429
material and it is referred to as thermal stress.
Chapter1
Stress and Strain
D E 'T
Vt x
S K Mondalâ€™s
Where, E = Modulus of elasticity
Thermal stress produces the same effect in the material similar to that of mechanical stress. A compressive stress will produce in the material with increase in temperature and the stress developed is tensile stress with decrease in temperature.
Let us take an example: A rod consists of two parts that are made of steel and copper as shown in figure below. The elastic modulus and coefficient of thermal expansion for steel are 200 GPa and 11.7 Ă— 106 per Â°C respectively and for copper 70 GPa and 21.6 Ă— 106 per Â°C respectively. If the temperature of the rod is raised by 50Â°C, determine the forces and stresses acting on the rod.
Answer: If we allow this rod to freely expand then free expansion
D 'T L
GT
11.7 u10 u 50 u 500 21.6 u10 u 50 u 750 6
6
1.1025 mm Compressive
But according to diagram only free expansion is 0.4 mm. Therefore restrained deflection of rod =1.1025 mm â€“ 0.4 mm = 0.7025 mm Let us assume the force required to make their elongation vanish be P which is the reaction force at the ends.
G
Â§ PL Âˇ Â¨ Â¸ ÂŠ AE ÂšSteel
Â§ PL Âˇ Â¨ Â¸ ÂŠ AE ÂšCu
P u 500 P u 750 2Â˝ 2Â˝ ÂS ÂS 9 9 ÂŽ u 0.075 Âž u 200 u 10 ÂŽ u 0.050 Âž u 70 u 10
4 4 ÂŻ Âż ÂŻ Âż or P 116.6 kN or 0.7025
Therefore, compressive stress on steel rod
V Steel
116.6 u 103
P ASteel
S 4
u 0.075
N/m2
26.39 MPa
2
And compressive stress on copper rod
V Cu
P ACu
116.6 u 103
S 4
u 0.050
N/m2
59.38 MPa
2
Page 13 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
1.17 Thermal stress on Brass and Mild steel combination A brass rod placed within a steel tube of exactly same length. The assembly is making in such a way that elongation of the combination will be same. To calculate the stress induced in the brass rod, steel tube when the combination is raised by toC then the following analogy have to do.
(a) Original bar before heating.
(b) Expanded position if the members are allowed to expand freely and independently after heating.
(c) Expanded position of the compound bar i.e. final position after heating.
x
Compatibility Equation:
G
x
G st G sf
G Bt G Bf
Equilibrium Equation:
V s As
V B AB
Assumption:
1. L = Ls
LB
2. D b ! D s 3. Steel Tension Brass Compression
Where, G = Expansion of the compound bar = AD in the above figure.
G st = Free expansion of the steel tube due to temperature rise toC = D s L t = AB in the above figure.
G sf = Expansion of the steel tube due to internal force developed by the unequal expansion. = BD in the above figure.
G Bt = Free expansion of the brass rod due to temperature rise toC = D b L t = AC in the above figure.
G Bf = Compression of the brass rod due to internal force developed by the unequal expansion. = BD in the above figure. And in the equilibrium equation Tensile force in the steel tube = Compressive force in the brass rod Where, V s = Tensile stress developed in the steel tube.
V B = Compressive stress developed in the brass rod. As = Cross section area of the steel tube. AB = Cross section area of the brass rod.
Let us take an example: See the Conventional Question Answer section of this chapter and the of 429 question is â€œConventional Question IES2008â€?Page and14itâ€™s answer.
Chapter1
Stress and Strain
S K Mondalâ€™s
1.18 Maximum stress and elongation due to rotation UZ 2L2
(i) V max
8
(ii) V max
UZ 2L2 2
and G L
and G L
UZ 2L3 12E
UZ 2L3 3E
For remember: You will get (ii) by multiplying by 4 of (i)
1.18 Creep When a member is subjected to a constant load over a long period of time it undergoes a slow permanent deformation and this is termed as â€œcreepâ€?. This is dependent on temperature. Usually at elevated temperatures creep is high.
x
The materials have its own different melting point; each will creep when the homologous temperature > 0.5. Homologous temp =
Testing temperature > 0.5 Melting temperature
A typical creep curve shows three distinct stages with different creep rates. After an initial rapid elongation Ç†o, the creep rate decrease with time until reaching the steady state. 1) Primary creep is a period of transient creep. The creep resistance of the material increases due to material deformation. 2) Secondary creep provides a nearly constant creep rate. The average value of the creep rate during this period is called the minimum creep rate. A stage of balance between competing. Strain hardening and recovery (softening) of the material. 3) Tertiary creep shows a rapid increase in the creep rate due to effectively reduced crosssectional area of the specimen leading to creep rupture or failure. In this stage intergranular cracking and/or formation of voids and cavities occur. Creep rate =c1 V
c2
Creep strain at any time = zero time strain intercept + creep rate Ă—Time = Â?0 Where,
c1 , c2 are constants V
c1 V c2 u t
stress
1.19 If a load P is applied suddenly to a bar then the stress & strain induced will be double than those obtained by an equal load applied gradually. Page 15 of 429
1.20 Stress produced by a load P in falling from height â€™hâ€™
Chapter1
Vd
Stress and Strain
ÂŞ
V ÂŤ1 1
S K Mondalâ€™s
2h Âş Â˝Â° ÂťÂž Äą, Â? L ÂťÂź Â°Âż
ÂŤÂŹ Â? being stress & strain produced by static load P & L=length of bar.
AÂŞ 2 AEh Âş ÂŤ1 1 Âť PÂŹ PL Âź
1.21 Loads shared by the materials of a compound bar made of bars x & y due to load W, Px
W.
Py
W.
1.22 Elongation of a compound bar, G
Ax Ex Ax Ex Ay E y Ay E y Ax Ex Ay E y
PL Ax Ex Ay E y
1.23 Tension Test
i)
True elastic limit: based on microstrain measurement at strains on order of 2 Ă— 106. Very low value and is related to the motion of a few hundred dislocations.
ii) Proportional limit: the highest stress at which stress is directly proportional to strain. iii) Elastic limit: is the greatest stress the material can withstand without any measurable permanent strain after unloading. Elastic limit > proportional limit.
iv) Yield strength is the stress required to produce a small specific amount of deformation. The offset yield strength can be determined by the stress corresponding to the intersection of the stressstrain curve and a line parallel to the elastic line offset by a strain of 0.2 or 0.1%. ( H = 0.002 or 0.001).
Page 16 of 429
Chapter1
x
Stress and Strain
S K Mondal’s
The offset yield stress is referred to proof stress either at 0.1 or 0.5% strain used for design and specification purposes to avoid the practical difficulties of measuring the elastic limit or proportional limit.
v) Tensile strength or ultimate tensile strength (UTS) V u is the maximum load Pmax divided by the original crosssectional area Ao of the specimen.
vi) % Elongation,
Lf Lo , is chiefly influenced by uniform elongation, which is dependent on the Lo
strainhardening capacity of the material.
vii) Reduction of Area: q
x
Ao Af Ao
Reduction of area is more a measure of the deformation required to produce failure and its chief contribution results from the necking process.
x
Because of the complicated state of stress state in the neck, values of reduction of area are dependent on specimen geometry, and deformation behaviour, and they should not be taken as true material properties.
x
RA is the most structuresensitive ductility parameter and is useful in detecting quality changes in the materials.
viii) Stressstrain response
1.24 Elastic strain and Plastic strain The strain present in the material after unloading is called the residual strain or plastic strain and the strain disappears during unloading is termed as recoverable or elastic strain. Equation of the straight line CB is given by
V total uE Plastic uE Elastic uE Carefully observe the following figures and understand which one is Elastic strain and which one is Plastic strain Page 17 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
Let us take an example: A 10 mm diameter tensile specimen has a 50 mm gauge length. The load corresponding to the 0.2% offset is 55 kN and the maximum load is 70 kN. Fracture occurs at 60 kN. The diameter after fracture is 8 mm and the gauge length at fracture is 65 mm. Calculate the following properties of the material from the tension test. (i)
% Elongation
(ii)
Reduction of Area (RA) %
(iii) Tensile strength or ultimate tensile strength (UTS) (iv) Yield strength (v)
Fracture strength
(vi) If E = 200 GPa, the elastic recoverable strain at maximum load (vii) If the elongation at maximum load (the uniform elongation) is 20%, what is the plastic strain at maximum load?
Answer: Given, Original area A0
Area at fracture Af
S
S 4
2
u 0.010 m2 2
4
u 0.008 m2
7.854 u 105 m2
5.027 u 105 m2
Original gauge length (L0) = 50 mm Gauge length at fracture (L) = 65 mm Therefore
(i) % Elongation
L L0 u 100% L0
(ii) Reduction of area (RA) = q
65 50 u 100 30% 50 A0 Af u 100% A0
7.854 5.027 u 100% 7.854
(iii) Tensile strength or Ultimate tensile strength (UTS), V u (iv) Yield strength V y
Py Ao
(v) Fracture strength V F
55 u 103 N/m2 5 7.854 u 10 PFracture Ao
Pmax Ao
36%
70 u 103 N/m2 5 7.854 u 10
700 MPa
60 u 103 N/m2 5 7.854 u 10
764MPa
Pmax / Ao 891u 106 9 Page 18Eof 429 200 u 10
(vi) Elastic recoverable strain at maximum load H E
0.0045
891 MPa
Chapter1 (vii) Plastic strain H P H total H E
Stress and Strain
0.2000 0.0045 0.1955
1.25 Elasticity This is the property of a material to regain its original shape after deformation when the external forces are removed. When the material is in elastic region the strain disappears completely after removal of the load, The stressstrain relationship in elastic region need not be linear and can be nonlinear (example rubber). The maximum stress value below which the strain is fully recoverable is called the elastic limit. It is represented by point A in figure. All materials are elastic to some extent but the degree varies, for example, both mild steel and rubber are elastic materials but steel is more elastic than rubber.
1.26 Plasticity When the stress in the material exceeds the elastic limit, the material enters into plastic phase where the strain can no longer be completely removed. Under plastic conditions materials ideally deform without any increase in stress. A typical stress strain diagram for an elasticperfectly plastic material is shown in the figure. MisesHenky criterion gives a good starting point for plasticity analysis.
1.27 Strain hardening If the material is reloaded from point C, it will follow the previous unloading path and line CB becomes its new elastic region with elastic limit defined by point B. Though the new elastic region CB resembles that of the initial elastic region OA, the internal structure of the material in the new state has changed. The change in the microstructure of the material is clear from the fact that the ductility of the material has come down due to strain hardening. When the material is reloaded, it follows the same path as that of a virgin material and fails on reaching the ultimate strength which remains unaltered due to the intermediate loading and unloading process.
Page 19 of 429
S K Mondalâ€™s
Chapter1
Stress and Strain
S K Mondal’s
1.28 Stress reversal and stressstrain hysteresis loop We know that fatigue failure begins at a local discontinuity and when the stress at the discontinuity exceeds elastic limit there is plastic strain. The cyclic plastic strain results crack propagation and fracture. When we plot the experimental data with reversed loading and the true stress strain hysteresis loops is found as shown below.
True stressstrain plot with a number of stress reversals Due to cyclic strain the elastic limit increases for annealed steel and decreases for cold drawn steel. Here the stress range is Ʀǔ. Ʀǆp and Ʀǆe are the plastic and elastic strain ranges, the total strain range being Ʀǆ. Considering that the total strain amplitude can be given as Ʀǆ = Ʀǆp+ Ʀǆe
Page 20 of 429
Chapter1
Stress and Strain
S K Mondal’s
OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20Years GATE Questions Stress in a bar due to selfweight GATE1.
Two identical circular rods of same diameter and same length are subjected to same magnitude of axial tensile force. One of the rods is made out of mild steel having the modulus of elasticity of 206 GPa. The other rod is made out of cast iron having the modulus of elasticity of 100 GPa. Assume both the materials to be homogeneous and isotropic and the axial force causes the same amount of uniform stress in both the rods. The stresses developed are within the proportional limit of the respective materials. Which of the following observations is correct? [GATE2003] (a) Both rods elongate by the same amount (b) Mild steel rod elongates more than the cast iron rod (c) Cast iron rod elongates more than the mild steel rod (d) As the stresses are equal strains are also equal in both the rods PL 1 or G L f [AsP, L and A is same] GATE1. Ans. (c) G L AE E G L mild steel ECI 100 ? G L CI ! G L MS EMS 206 G L C.I GATE2.
A steel bar of 40 mm × 40 mm square crosssection is subjected to an axial compressive load of 200 kN. If the length of the bar is 2 m and E = 200 GPa, the elongation of the bar will be: [GATE2006] (a) 1.25 mm (b) 2.70 mm (c) 4.05 mm (d) 5.40 mm
GATE2. Ans. (a) G L
PL AE
200 u 1000 u 2 0.04 u 0.04 u 200 u 109
m 1.25mm
True stress and true strain GATE3.
The ultimate tensile strength of a material is 400 MPa and the elongation up to maximum load is 35%. If the material obeys power law of hardening, then the true stresstrue strain relation (stress in MPa) in the plastic deformation range is: [GATE2006] 0.30 0.30 0.35 (b) V 775H (c) V 540H (d) V 775H 0.35 (a) V 540H GATE3. Ans. (c) A true stress – true strain curve in tension V kH n k = Strength coefficient = 400 × (1.35) = 540 MPa n = Strain – hardening exponent = 0.35
Elasticity and Plasticity GATE4.
An axial residual compressive stress due to a manufacturing process is present on the outer surface of a rotating shaft subjected to bending. Under a given Page 21 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
bending load, the fatigue life of the shaft in the presence of the residual compressive stress is: [GATE2008] (a) Decreased (b) Increased or decreased, depending on the external bending load (c) Neither decreased nor increased (d) Increased GATE4. Ans. (d)
A cantileverloaded rotating beam, showing the normal distribution of surface stresses. (i.e., tension at the top and compression at the bottom)
The residual compressive stresses induced.
Net stress pattern obtained when loading a surface treated beam. The reduced magnitude of the tensile stresses contributes to increased fatigue life. GATE5.
A static load is mounted at the centre of a shaft rotating at uniform angular velocity. This shaft will be designed for [GATE2002] (a) The maximum compressive stress (static) (b) The maximum tensile stress (static) (c) The maximum bending moment (static) (d) Fatigue loading GATE5. Ans. (d) GATE6.
Fatigue strength of a rod subjected to cyclic axial force is less than that of a rotating beam of the same dimensions subjected to steady lateral force because (a) Axial stiffness is less than bending stiffness [GATE1992] (b) Of absence of centrifugal effects in the rod (c) The number of discontinuities vulnerable to fatigue are more in the rod (d) At a particular time the rod has only one type of stress whereas the beam has both the tensile and compressive stresses. GATE6. Ans. (d)
Relation between the Elastic Modulii GATE7.
A rod of length L and diameter D is subjected to a tensile load P. Which of the following is sufficient to calculate the resulting change in diameter? (a) Young's modulus (b) Shear modulus [GATE2008] (c) Poisson's ratio (d) Both Young's modulus and shear modulus Page 22 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
GATE7. Ans. (d) For longitudinal strain we need Young's modulus and for calculating transverse strain we need Poisson's ratio. We may calculate Poisson's ratio from E 2G (1 P ) for that we need Shear modulus. GATE8.
In terms of Poisson's ratio (Îź) the ratio of Young's Modulus (E) to Shear Modulus (G) of elastic materials is [GATE2004]
(a) 2(1 P )
1 (c) (1 P ) 2
(b) 2(1 P )
(d )
1 (1 P ) 2
GATE8. Ans. (a) GATE9.
The relationship between Young's modulus (E), Bulk modulus (K) and Poisson's ratio (Îź) is given by: [GATE2002] (b) K (a) E 3 K 1 2P
3 E 1 2P
(c) E
3 K 1 P
GATE9. Ans. (a) Remember
(d) K
E
2G 1 P
3K 1 2P
3 E 1 P
9KG 3K G
Stresses in compound strut GATE10. In a bolted joint two members are connected with an axial tightening force of 2200 N. If the bolt used has metric threads of 4 mm pitch, then torque required for achieving the tightening force is (a) 0.7Nm (b) 1.0 Nm (c) 1.4Nm (d) 2.8Nm GATE10. Ans. (c) T
Fur
2200 u
0.004 Nm 2S
[GATE2004] 1.4Nm
GATE11. The figure below shows a steel rod of 25 mm2 cross sectional area. It is loaded at four points, K, L, M and N. [GATE2004, IES 1995, 1997, 1998]
Assume Esteel = 200 GPa. The total change in length of the rod due to loading is: (a) 1 Îźm (b) 10 Îźm (c) 16 Îźm (d) 20 Îźm
GATE11. Ans. (b) First draw FBD of all parts separately then
Total change in length =
PL
Âœ AE
GATE12. A bar having a crosssectional area of 700mm2 is subjected to axial loads at the positions indicated. The value of stress in the segment QR is: [GATE2006] Page 23 of 429
Chapter1
Stress and Strain
P (a) 40 MPa GATE12. Ans. (a)
Q (b) 50 MPa
S K Mondal’s
R (c) 70 MPa
S (d) 120 MPa
F.B.D
V QR
P A
28000 MPa 700
40MPa
GATE13. An ejector mechanism consists of a helical compression spring having a spring constant of K = 981 × 103 N/m. It is precompressed by 100 mm from its free state. If it is used to eject a mass of 100 kg held on it, the mass will move up through a distance of (a) 100mm (b) 500mm (c) 981 mm (d) 1000mm [GATE2004] GATE13. Ans. (a) No calculation needed it is precompressed by 100 mm from its free state. So it can’t move more than 100 mm. choice (b), (c) and (d) out. GATE14. The figure shows a pair of pinjointed grippertongs holding an object weighing 2000 N. The coefficient of friction (μ) at the gripping surface is 0.1 XX is the line of action of the input force and YY is the line of application of gripping force. If the pinjoint is assumed to be frictionless, then magnitude of force F required to hold the weight is: (a) 1000 N (b) 2000 N (c) 2500 N (d) 5000 N
[GATE2004] GATE14. Ans. (d) Frictional force required = 2000 N Force needed to produce 2000N frictional force at YY section = So for each side we need (Fy) = 10000 N force
Page 24 of 429
2000 0.1
20000N
Chapter1
Stress and Strain
Taking moment about PIN Fy u 50 Fy u 50 F u 100 or F 100
10000 u 50 100
S K Mondalâ€™s
5000N
GATE15. A uniform, slender cylindrical rod is made of a homogeneous and isotropic material. The rod rests on a frictionless surface. The rod is heated uniformly. If the radial and longitudinal thermal stresses are represented by Ç”r and Ç”z, respectively, then [GATE2005]
(a) V r
0, V z
0
(b) V r z 0, V z
0
(c ) V r
0, V z z 0
(d ) V r z 0, V z z 0
GATE15. Ans. (a) Thermal stress will develop only when you prevent the material to contrast/elongate. As here it is free no thermal stress will develop.
Tensile Test GATE16. A test specimen is stressed slightly beyond the yield point and then unloaded. Its yield strength will [GATE1995] (a) Decrease (b) Increase (c) Remains same (d) Becomes equal to ultimate tensile strength GATE16. Ans. (b)
GATE17. Under repeated loading a material has the stressstrain curve shown in figure, which of the following statements is true? (a) The smaller the shaded area, the better the material damping (b) The larger the shaded area, the better the material damping (c) Material damping is an independent material property and does not depend on this curve (d) None of these GATE17. Ans. (a)
[GATE1999]
Previous 20Years IES Questions Stress in a bar due to selfweight IES1.
A solid uniform metal bar of diameter D and length L is hanging vertically from its upper end. The elongation of the bar due to self weight is: [IES2005] (a) Proportional to L and inversely proportional to D2 (b) Proportional to L2 and inversely proportional to D2 (c) Proportional of L but independent of D (d) Proportional of U but independent of D Page 25 of 429
Chapter1
Stress and Strain
IES1. Ans. (a) G
WL 2AE
WL 2u
S D2 4
?G u L uE
& Gu
S K Mondalâ€™s
1 D2
IES2.
The deformation of a bar under its own weight as compared to that when subjected to a direct axial load equal to its own weight will be: [IES1998] (a) The same (b) Onefourth (c) Half (d) Double IES2. Ans. (c) IES3.
A rigid beam of negligible weight is supported in a horizontal position by two rods of steel and aluminum, 2 m and 1 m long having values of cross  sectional areas 1 cm2 and 2 cm2 and E of 200 GPa and 100 GPa respectively. A load P is applied as shown in the figure [IES2002]
If the rigid beam is to remain horizontal then (a) The forces on both sides should be equal (b) The force on aluminum rod should be twice the force on steel (c) The force on the steel rod should be twice the force on aluminum (d) The force P must be applied at the centre of the beam IES3. Ans. (b)
Bar of uniform strength IES4.
Which one of the following statements is correct? [IES 2007] A beam is said to be of uniform strength, if (a) The bending moment is the same throughout the beam (b) The shear stress is the same throughout the beam (c) The deflection is the same throughout the beam (d) The bending stress is the same at every section along its longitudinal axis IES4. Ans. (d) IES5.
Which one of the following statements is correct? [IES2006] Beams of uniform strength vary in section such that (a) bending moment remains constant (b) deflection remains constant (c) maximum bending stress remains constant (d) shear force remains constant IES5. Ans. (c) IES6.
For bolts of uniform strength, the shank diameter is made equal to (a) Major diameter of threads (b) Pitch diameter of threads (c) Minor diameter of threads (d) Nominal diameter of threads IES6. Ans. (c)
[IES2003]
IES7.
A bolt of uniform strength can be developed by [IES1995] (a) Keeping the core diameter of threads equal to the diameter of unthreaded portion of the bolt Page 26 of 429 (b) Keeping the core diameter smaller than the diameter of the unthreaded portion
Chapter1
Stress and Strain (c)
S K Mondalâ€™s
Keeping the nominal diameter of threads equal the diameter of unthreaded portion of the bolt One end fixed and the other end free
(d) IES7. Ans. (a)
Elongation of a Taper Rod IES8.
Two tapering bars of the same material are subjected to a tensile load P. The lengths of both the bars are the same. The larger diameter of each of the bars is D. The diameter of the bar A at its smaller end is D/2 and that of the bar B is D/3. What is the ratio of elongation of the bar A to that of the bar B? [IES2006] (a) 3 : 2 (b) 2: 3 (c) 4 : 9 (d) 1: 3 PL IES8. Ans. (b) Elongation of a taper rod G l
S ddE 4 1 2 G l A d2 B Â§ D / 3 Âˇ 2 or G l B d2 A Â¨ÂŠ D / 2 Â¸Âš 3 IES9.
A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation? [IES2004] (a) 10% (b) 5% (c) 1% (d) 0.5% PL PL IES9. Ans. (c) Actual elongation of the bar G l act Â§S Âˇ Â§S Âˇ Â¨ 4 d1d2 Â¸ E Â¨ 4 u 1.1D u 0.9D Â¸ E ÂŠ Âš ÂŠ Âš PL Calculated elongation of the bar G l Cal S D2 uE 4 G l act G l cal Â§ Âˇ D2 ? Error %
u 100 Â¨ 1Â¸ u 100% 1% G l cal ÂŠ 1.1D u 0.9D Âš
IES10.
The stretch in a steel rod of circular section, having a length 'l' subjected to a tensile load' P' and tapering uniformly from a diameter d1 at one end to a [IES1995] diameter d2 at the other end, is given (a)
Pl 4 Ed1d 2
(b)
pl.S Ed1d 2
IES10. Ans. (d) Actual elongation of the bar G l act
IES11.
(c)
pl.S 4 Ed1d 2
(d)
4 pl S Ed1d 2
PL Â§S
Âˇ Â¨ d1d2 Â¸ E ÂŠ4 Âš
A tapering bar (diameters of end sections being d1 and d2 a bar of uniform crosssection â€™dâ€™ have the same length and are subjected the same axial pull. Both the bars will have the same extension ifâ€™dâ€™ is equal to [IES1998]
a
d1 d 2 2
b
d1d 2
c
d1d 2 2
d
d1 d 2 2
IES11. Ans. (b)
Poissonâ€™s ratio IES12.
In the case of an engineering material under unidirectional stress in the xdirection, the Poisson's ratio is equal to (symbols have the usual meanings) [IAS 1994, IES2000] Page 27 of 429
Chapter1
Stress and Strain (a)
Hy Hx
(b)
Hy Vx
S K Mondalâ€™s (c)
Vy Vx
(d)
Vy Hx
IES12. Ans. (a) IES13.
Which one of the following is correct in respect of Poisson's ratio (v) limits for an isotropic elastic solid? [IES2004] (b) 1/ 4 dQ d1/ 3 (c) 1dQ d1/ 2 (d) 1/ 2 dQ d1/ 2 (a) f dQ d f IES13. Ans. (c) Theoretically 1 P 1/ 2 but practically 0 P 1/ 2
IES14.
Match ListI (Elastic properties of an isotropic elastic material) with ListII (Nature of strain produced) and select the correct answer using the codes given below the Lists: [IES1997] ListI ListII A. Young's modulus 1. Shear strain B. Modulus of rigidity 2. Normal strain C. Bulk modulus 3. Transverse strain D. Poisson's ratio 4. Volumetric strain Codes: A B C D A B C D (a) 1 2 3 4 (b) 2 1 3 4 (c) 2 1 4 3 (d) 1 2 4 3 IES14. Ans. (c) IES15.
If the value of Poisson's ratio is zero, then it means that (a) The material is rigid. (b) The material is perfectly plastic. (c) There is no longitudinal strain in the material (d) The longitudinal strain in the material is infinite. IES15. Ans. (a) If Poisson's ratio is zero, then material is rigid.
[IES1994]
IES16.
[IES1992]
Which of the following is true (Îź= Poisson's ratio) (b) 1 P 0 (c) 1 P 1 (a) 0 P 1/ 2 IES16. Ans. (a)
(d) f
P f
Elasticity and Plasticity IES17.
If the area of crosssection of a wire is circular and if the radius of this circle decreases to half its original value due to the stretch of the wire by a load, then the modulus of elasticity of the wire be: [IES1993] (a) Onefourth of its original value (b) Halved (c) Doubled (d) Unaffected IES17. Ans. (d) Note: Modulus of elasticity is the property of material. It will remain same. IES18.
The relationship between the Lameâ€™s constant â€˜ÇŒâ€™, Youngâ€™s modulus â€˜Eâ€™ and the Poissonâ€™s ratio â€˜Ç?â€™ [IES1997]
a O
EP 1 P 1 2P
(b)O
EP 1 2P 1 P
c O
EP 1 P
d O
EP 1 P
IES18. Ans. (a) IES19.
Which of the following pairs are correctly matched? [IES1994] 1. Resilienceâ€Śâ€Śâ€Śâ€Śâ€Ś Resistance to deformation. 2. Malleability â€Śâ€Śâ€Śâ€Ś..Shape change. 3. Creep ........................ Progressive deformation. 4. Plasticity .... â€Śâ€Śâ€Śâ€Ś.Permanent deformation. Select the correct answer using the codes given below: Codes: (a) 2, 3 and 4 (b) 1, 2 and 3 (c) 1, 2 and 4 (d) 1, 3 and 4 IES19. Ans. (a) Strain energy stored by a body within elastic limit is known as resilience. Page 28 of 429
Chapter1
Stress and Strain
S K Mondal’s
Creep and fatigue IES20.
What is the phenomenon of progressive extension of the material i.e., strain increasing with the time at a constant load, called? [IES 2007] (a) Plasticity (b) Yielding (b) Creeping (d) Breaking IES20. Ans. (c) IES21.
The correct sequence of creep deformation in a creep curve in order of their elongation is: [IES2001] (a) Steady state, transient, accelerated (b) Transient, steady state, accelerated (c) Transient, accelerated, steady state (d) Accelerated, steady state, transient IES21. Ans. (b) IES22.
The highest stress that a material can withstand for a specified length of time without excessive deformation is called [IES1997] (a) Fatigue strength (b) Endurance strength (c) Creep strength (d) Creep rupture strength IES22. Ans. (c) IES23.
Which one of the following features improves the fatigue strength of a metallic material? [IES2000] (a) Increasing the temperature (b) Scratching the surface (c) Overstressing (d) Under stressing IES23. Ans. (d) IES24.
Consider the following statements: [IES1993] For increasing the fatigue strength of welded joints it is necessary to employ 1. Grinding 2. Coating 3. Hammer peening Of the above statements (a) 1 and 2 are correct (b) 2 and 3 are correct (c) 1 and 3 are correct (d) 1, 2 and 3 are correct IES24. Ans. (c) A polished surface by grinding can take more number of cycles than a part with rough surface. In Hammer peening residual compressive stress lower the peak tensile stress
Relation between the Elastic Modulii IES25.
For a linearly elastic, isotropic and homogeneous material, the number of elastic constants required to relate stress and strain is: [IAS 1994; IES1998] (a) Two (b) Three (c) Four (d) Six IES25. Ans. (a) E, G, K and Ǎ represent the elastic modulus, shear modulus, bulk modulus and Poisson's ratio respectively of a linearly elastic, isotropic and homogeneous material. To express the stressstrain relations completely for this material, at least [IES2006] (a) E, G and Ǎ must be known (b) E, K and Ǎ must be known (c) Any two of the four must be known (d) All the four must be known IES26. Ans. (c) IES27. The number of elastic constants for a completely anisotropic elastic material which follows Hooke's law is: [IES1999] (a) 3 (b) 4 (c) 21 (d) 25 IES27. Ans. (c)
IES26.
IES28.
What are the materials which show direction dependent properties, called? (a) Homogeneous materials (b) Viscoelastic materials [IES 2007] (c) Isotropic materials (d) Anisotropic materials IES28. Ans. (d) IES29.
Page 29 of 429
An orthotropic material, under plane stress condition will have:
[IES2006]
Chapter1
Stress and Strain
(a) 15 independent elastic constants (c) 5 independent elastic constants IES29. Ans. (d)
S K Mondalâ€™s
(b) 4 independent elastic constants (d) 9 independent elastic constants
IES30.
Match ListI (Properties) with ListII (Units) and select the correct answer using the codes given below the lists: [IES2001] List I List II A. Dynamic viscosity 1. Pa B. Kinematic viscosity 2. m2/s C. Torsional stiffness 3. Ns/m2 D. Modulus of rigidity 4. N/m Codes: A B C D A B C D (a) 3 2 4 1 (b) 5 2 4 3 (b) 3 4 2 3 (d) 5 4 2 1 IES30. Ans. (a) IES31.
Young's modulus of elasticity and Poisson's ratio of a material are 1.25 Ă— 105 MPa and 0.34 respectively. The modulus of rigidity of the material is: [IAS 1994, IES1995, 2001, 2002, 2007] (b) 0.4664 Ă— 105 Mpa (a) 0.4025 Ă—105 Mpa (d) 0.9469 Ă— 105 MPa (c) 0.8375 Ă— 105 MPa 5 IES31. Ans.(b) E 2G (1 P ) or 1.25x10 = 2G(1+0.34) or G = 0.4664 Ă— 105 MPa
IES32.
In a homogenous, isotropic elastic material, the modulus of elasticity E in terms of G and K is equal to [IAS1995, IES  1992] (a)
G 3K 9 KG
(b)
3G K 9 KG
(c)
9 KG G 3K
(d)
9 KG K 3G
IES32. Ans. (c) IES33.
What is the relationship between the linear elastic properties Young's modulus (E), rigidity modulus (G) and bulk modulus (K)? [IES2008] 1 9 3 3 9 1 9 3 1 9 1 3 (a) (b) (c) (d) E K G E K G E K G E K G 9KG IES33. Ans. (d) E 2G 1 P 3K 1 2 P
3K G
IES34.
What is the relationship between the liner elastic properties Youngâ€™s modulus (E), rigidity modulus (G) and bulk modulus (K)? [IES2009] (a)
E
IES34. Ans. (d) E
KG 9K G 2G 1 P
(b)
E
9KG K G
3K 1 2P
(c)
E
9 KG K 3G
(d)
E
9 KG 3K G
9KG 3K G
IES35.
If E, G and K denote Young's modulus, Modulus of rigidity and Bulk Modulus, respectively, for an elastic material, then which one of the following can be possibly true? [IES2005] (a) G = 2K (b) G = E (c) K = E (d) G = K = E 9KG IES35. Ans.(c) E 2G 1 P 3K 1 2P
3K G 1 the value of P must be between 0 to 0.5 so E never equal to G but if P then 3 E k so ans. is c IES36.
If a material had a modulus of elasticity of 2.1 Ă— 106 kgf/cm2 and a modulus of rigidity of 0.8 Ă— 106 kgf/cm2 then the approximate value of the Poisson's ratio of the material would be: [IES1993] Page 30 of 429 (a) 0.26 (b) 0.31 (c) 0.47 (d) 0.5
Chapter1
Stress and Strain
IES36. Ans. (b) Use E
S K Mondalâ€™s
2G 1 P
The modulus of elasticity for a material is 200 GN/m2 and Poisson's ratio is 0.25. What is the modulus of rigidity? [IES2004] (b) 125 GN/m2 (c) 250 GN/m2 (d) 320 GN/m2 (a) 80 GN/m2 E 200 80GN / m2 IES37. Ans. (a) E 2G 1 P or G 2 1 P 2 u 1 0.25
IES37.
IES38.
Consider the following statements: [IES2009] 1. Twodimensional stresses applied to a thin plate in its own plane represent the plane stress condition. 2. Under plane stress condition, the strain in the direction perpendicular to the plane is zero. 3. Normal and shear stresses may occur simultaneously on a plane. Which of the above statements is /are correct? (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 3 IES38. Ans. (d) Under plane stress condition, the strain in the direction perpendicular to the plane is not zero. It has been found experimentally that when a body is stressed within elastic limit, the lateral strain bears a constant ratio to the linear strain. [IES2009]
Stresses in compound strut IES39.
Eight bolts are to be selected for fixing the cover plate of a cylinder subjected to a maximum load of 980Âˇ175 kN. If the design stress for the bolt material is [IES2008] 315 N/mm2, what is the diameter of each bolt? (a) 10 mm (b) 22 mm (c) 30 mm (d) 36 mm
IES39. Ans. (b) Total load P
IES40.
8 uV u
S d2 4
or d
P 2SV
980175 2S u 315
22.25 mm
For a composite consisting of a bar enclosed inside a tube of another material when compressed under a load 'w' as a whole through rigid collars at the end of the bar. The equation of compatibility is given by (suffixes 1 and 2) refer to bar and tube respectively [IES1998]
(a) W1 W2
W
(b) W1 W2
Const. (c)
W1 A1 E1
W2 A2 E2
(d )
W1 A1 E2
W2 A2 E1
IES40. Ans. (c) Compatibility equation insists that the change in length of the bar must be compatible with the boundary conditions. Here (a) is also correct but it is equilibrium equation. IES41.
When a composite unit consisting of a steel rod surrounded by a cast iron tube is subjected to an axial load. [IES2000] Assertion (A): The ratio of normal stresses induced in both the materials is equal to the ratio of Young's moduli of respective materials. Reason (R): The composite unit of these two materials is firmly fastened together at the ends to ensure equal deformation in both the materials. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES41. Ans. (a) IES42.
The figure below shows a steel rod of 25 mm2 cross sectional area. It is loaded at four points, K, L, M and N. [GATE2004, IES 1995, 1997, 1998]
Page 31 of 429
Chapter1
Stress and Strain
S K Mondal’s
Assume Esteel = 200 GPa. The total change in length of the rod due to loading is (a) 1 μm (b) 10 μm (c) 16 μm (d) 20 μm IES42. Ans. (b) First draw FBD of all parts separately then
Total change in length =
IES43.
PL
AE
The reactions at the rigid supports at A and B for the bar loaded as shown in the figure are respectively. (a) 20/3 kN,10/3 kN (b) 10/3 kN, 20/3 kN (c) 5 kN, 5 kN (d) 6 kN, 4 kN [IES2002;
IAS
2003] IES43. Ans. (a) Elongation in AC = length reduction in CB
R A u 1 RB u 2 AE AE And RA + RB = 10 IES44.
Which one of the following is correct? [IES2008] When a nut is tightened by placing a washer below it, the bolt will be subjected to (a) Compression only (b) Tension (c) Shear only (d) Compression and shear IES44. Ans. (b) IES45.
Which of the following stresses are associated with the tightening of nut on a bolt? [IES1998] 1. Tensile stress due to the stretching of bolt 2. Bending stress due to the bending of bolt 3. Crushing and shear stresses in threads 4. Torsional shear stress due to frictional resistance between the nut and the bolt. Select the correct answer using the codes given below Codes: (a) 1, 2 and 4 (b) 1, 2 and 3 (c) 2, 3 and 4 (d) 1, 3 and 4 IES45. Ans. (d)
Thermal effect IES46.
A 100 mm × 5 mm × 5 mm steel bar free to expand is heated from 15°C to 40°C. What shall be developed? [IES2008] (a) Tensile stress (b) Compressive stress (c) Shear stress (d) No stress IES46. Ans. (d) If we resist to expand then only stress will develop. IES47.
Which one of the following statements is correct? [GATE1995; IES 2007] Page 32 of 429 it will develop If a material expands freely due to heating,
Chapter1
Stress and Strain
(a) Thermal stress IES47. Ans. (d) IES48.
(b) Tensile stress
S K Mondalâ€™s
(c) Compressive stress
(d) No stress
A cube having each side of length a, is constrained in all directions and is heated uniformly so that the temperature is raised to TÂ°C. If Ç‚ is the thermal coefficient of expansion of the cube material and E the modulus of elasticity, the stress developed in the cube is: [IES2003] (a)
D TE J
(b)
D TE 1 2J
(c)
D TE 2J
(d)
D TE 1 2J
3
3 3 %V T p a 1 BT a IES48. Ans. (b) V K a3 P Or 3BT E 3 1 2H
IES49.
Consider the following statements: Thermal stress is induced in a component in general, when 1. A temperature gradient exists in the component 2. The component is free from any restraint 3. It is restrained to expand or contract freely Which of the above statements are correct? (a) 1 and 2 (b) 2 and 3 (c) 3 alone IES49. Ans. (c) IES50.
[IES2002]
(d) 2 alone
A steel rod 10 mm in diameter and 1m long is heated from 20Â°C to 120Â°C, E = 200 GPa and Ç‚ = 12 Ă— 106 per Â°C. If the rod is not free to expand, the thermal stress developed is: [IAS2003, IES1997, 2000, 2006] (a) 120 MPa (tensile) (b) 240 MPa (tensile) (c) 120 MPa (compressive) (d) 240 MPa (compressive)
IES50. Ans. (d) D E't
12 u 10 u 200 u 10 u 120 20
6
3
240MPa
It will be compressive as elongation restricted. IES51.
A cube with a side length of 1 cm is heated uniformly 1Â° C above the room temperature and all the sides are free to expand. What will be the increase in volume of the cube? (Given coefficient of thermal expansion is Ç‚ per Â°C) (b) 2 Ç‚ cm3 (c) Ç‚ cm3 (d) zero [IES2004] (a) 3 Ç‚ cm3 IES51. Ans. (a) coefficient of volume expansion J 3 u co efficient of linear exp ansion D
IES52.
A bar of copper and steel form a composite system. [IES2004] They are heated to a temperature of 40 Â° C. What type of stress is induced in the copper bar? (a) Tensile (b) Compressive (c) Both tensile and compressive (d) Shear IES52. Ans. (b) IES53.
ÄŽ =12.5Ă—106 / o C, E = 200GPa If the rod fitted strongly between the supports as shown in the figure, is heated, the stress induced in it due to 20oC rise in temperature will be: [IES1999] (a) 0.07945 MPa (b) 0.07945 MPa (c) 0.03972 MPa (d) 0.03972 MPa
Page 33 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
IES53. Ans. (b) Let compression of the spring = x m Therefore spring force = kx kN Expansion of the rod due to temperature rise = LD't Reduction in the length due to compression force = Now LD't Or x
kx u L AE
kx u L
x AE 0.5 u 12.5 u 106 u 20
0.125 mm Â Â˝ Â°Â° Â°Â° 50 u 0.5 ÂŽ1 Âž 2 Â° S u 0.010 u 200 u 106 Â° 4 ÂŻÂ° ÂżÂ° kx 50 u 0.125 ? Compressive stress = 0.07945MPa A Â§ S u 0.0102 Âˇ Â¨ Â¸ 4 ÂŠ Âš
IES54.
The temperature stress is a function of [IES1992] 1. Coefficient of linear expansion 2. Temperature rise 3. Modulus of elasticity The correct answer is: (a) 1 and 2 only (b) 1 and 3 only (c) 2 and 3 only (d) 1, 2 and 3 IES54. Ans. (d) Stress in the rod due to temperature rise = D't u E
Impact loading IES55.
Assertion (A): Ductile materials generally absorb more impact loading than a brittle material [IES2004] Reason (R): Ductile materials generally have higher ultimate strength than brittle materials (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES55. Ans. (c) IES56.
Assertion (A): Specimens for impact testing are never notched. [IES1999] Reason (R): A notch introduces triaxial tensile stresses which cause brittle fracture. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES56. Ans. (d) A is false but R is correct. Page 34 of 429
Chapter1
Stress and Strain
S K Mondal’s
Tensile Test IES57.
During tensiletesting of a specimen using a Universal Testing Machine, the parameters actually measured include [IES1996] (a) True stress and true strain (b) Poisson’s ratio and Young's modulus (c) Engineering stress and engineering strain (d) Load and elongation IES57. Ans. (d) IES58.
In a tensile test, near the elastic limit zone (a) Tensile stress increases at a faster rate (b) Tensile stress decreases at a faster rate (c) Tensile stress increases in linear proportion to the stress (d) Tensile stress decreases in linear proportion to the stress IES58. Ans. (b)
[IES2006]
IES59.
Match ListI (Types of Tests and Materials) with ListII (Types of Fractures) and select the correct answer using the codes given below the lists: List I ListII [IES2002; IAS2004] (Types of Tests and Materials) (Types of Fractures) A. Tensile test on CI 1. Plain fracture on a transverse plane B. Torsion test on MS 2. Granular helecoidal fracture C. Tensile test on MS 3. Plain granular at 45° to the axis D. Torsion test on CI 4. Cup and Cone 5. Granular fracture on a transverse plane Codes: A B C D A B C D (a) 4 2 3 1 (c) 4 1 3 2 (b) 5 1 4 2 (d) 5 2 4 1 IES59. Ans. (d) IES60.
Which of the following materials generally exhibits a yield point? [IES2003] (a) Cast iron (b) Annealed and hotrolled mild steel (c) Soft brass (d) Coldrolled steel IES60. Ans. (b) IES61.
For most brittle materials, the ultimate strength in compression is much large then the ultimate strength in tension. The is mainly due to [IES1992] (a) Presence of flaws and microscopic cracks or cavities (b) Necking in tension (c) Severity of tensile stress as compared to compressive stress (d) Nonlinearity of stressstrain diagram IES61. Ans. (a) IES62.
What is the safe static tensile load for a M36 × 4C bolt of mild steel having yield stress of 280 MPa and a factor of safety 1.5? [IES2005] (a) 285 kN (b) 190 kN (c) 142.5 kN (d) 95 kN W S d2 or W V u ; IES62. Ans. (b) V c c 4 S d2 4 W V c u S u d2 280 u S u 362 Wsafe N 190kN fos fos u 4 1.5 u 4
IES63.
Which one of the following properties is more sensitive to increase in strain rate? [IES2000] (a) Yield strength (b) Proportional limit (c) Elastic limit (d) Tensile strength IES63. Ans. (b) Page 35 of 429
Chapter1
Stress and Strain
S K Mondal’s
IES64.
A steel hub of 100 mm internal diameter and uniform thickness of 10 mm was heated to a temperature of 300oC to shrinkfit it on a shaft. On cooling, a crack developed parallel to the direction of the length of the hub. Consider the following factors in this regard: [IES1994] 1. Tensile hoop stress 2. Tensile radial stress 3. Compressive hoop stress 4. Compressive radial stress The cause of failure is attributable to (a) 1 alone (b) 1 and 3 (c) 1, 2 and 4 (d) 2, 3 and 4 IES64. Ans. (a) A crack parallel to the direction of length of hub means the failure was due to tensile hoop stress only. IES65.
If failure in shear along 45° planes is to be avoided, then a material subjected to uniaxial tension should have its shear strength equal to at least [IES1994] (a) Tensile strength (b) Compressive strength (c) Half the difference between the tensile and compressive strengths. (d) Half the tensile strength. IES65. Ans. (d) IES66. Select the proper sequence [IES1992] 1. Proportional Limit 2. Elastic limit 3. Yielding 4. Failure (a) 2, 3, 1, 4 (b) 2, 1, 3, 4 (c) 1, 3, 2, 4 (d) 1, 2, 3, 4 IES66. Ans. (d)
Previous 20Years IAS Questions Stress in a bar due to selfweight IAS1.
A heavy uniform rod of length 'L' and material density 'ǅ' is hung vertically with its top end rigidly fixed. How is the total elongation of the bar under its own weight expressed? [IAS2007]
2G L2 g E
(a)
(b)
G L2 g E
IAS1. Ans. (d) Elongation due to self weight = IAS2.
(c)
WL 2 AE
G L2 g
(d)
2E G ALg L G L2 g 2 AE 2E
G L2 g 2E
A rod of length 'l' and crosssection area ‘A’ rotates about an axis passing through one end of the rod. The extension produced in the rod due to centrifugal forces is (w is the weight of the rod per unit length and Z is the angular velocity of rotation of the rod). [IAS 1994] (a)
Zwl 2 gE
(b)
Z 2 wl 3 3 gE
(c)
Z 2 wl 3 gE
IAS2. Ans. (b)
Page 36 of 429
(d)
3 gE Z 2 wl 3
Chapter1
Stress and Strain
S K Mondal’s
Elongation of a Taper Rod IAS3.
A rod of length, " L " tapers uniformly from a diameter ''D1' to a diameter ''D2' and carries an axial tensile load of "P". The extension of the rod is (E represents the modulus of elasticity of the material of the rod) [IAS1996] (a)
4 P1 S ED1 D2
(b)
4 PE1 S D1 D2
(c)
IAS3. Ans. (a) The extension of the taper rod =
S EP1 4 D1D2
(d)
S P1 4 ED1D2
Pl §S · ¨ 4 D1D2 ¸ .E © ¹
Poisson’s ratio IAS4.
In the case of an engineering material under unidirectional stress in the xdirection, the Poisson's ratio is equal to (symbols have the usual meanings) [IAS 1994, IES2000] (a)
Hy Hx
(b)
Hy Vx
(c)
Vy Vx
(d)
Vy Hx
IAS4. Ans. (a) IAS5.
Assertion (A): Poisson's ratio of a material is a measure of its ductility. Reason (R): For every linear strain in the direction of force, Poisson's ratio of the material gives the lateral strain in directions perpendicular to the direction of force. [IAS1999] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS5. ans. (d) IAS6.
Assertion (A): Poisson's ratio is a measure of the lateral strain in all direction perpendicular to and in terms of the linear strain. [IAS1997] Reason (R): The nature of lateral strain in a uniaxially loaded bar is opposite to that of the linear strain. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS6. Ans. (b)
Elasticity and Plasticity IAS7.
A weight falls on a plunger fitted in a container filled with oil thereby producing a pressure of 1.5 N/mm2 in the oil. The Bulk Modulus of oil is 2800 N/mm2. Given this situation, the volumetric compressive strain produced in the oil will be: [IAS1997] (b) 800 × 106 (c) 268 × 106 (d) 535 × 106 (a) 400 × 106 P P 1.5 IAS7. Ans. (d) Bulk modulus of elasticity (K) = or H v 535 u 106 Hv K 2800
Relation between the Elastic Modulii IAS8.
For a linearly elastic, isotropic and homogeneous material, the number of elastic constants required to relate stress and strain is: [IAS 1994; IES1998] (a) Two (b) Three (c) Four (d) Six IAS8. Ans. (a) Page 37 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
IAS9.
The independent elastic constants for a homogenous and isotropic material are (a) E, G, K, v (b) E, G, K (c) E, G, v (d) E, G [IAS1995] IAS9. Ans. (d) IAS10.
The unit of elastic modulus is the same as those of [IAS 1994] (a) Stress, shear modulus and pressure (b) Strain, shear modulus and force (c) Shear modulus, stress and force (d) Stress, strain and pressure. IAS10. Ans. (a) IAS11.
Young's modulus of elasticity and Poisson's ratio of a material are 1.25 Ă— 105 MPa and 0.34 respectively. The modulus of rigidity of the material is: [IAS 1994, IES1995, 2001, 2002, 2007] (b) 0.4664 Ă— 105 MPa (a) 0.4025 Ă— 105 MPa 5 (d) 0.9469 Ă— 105 MPa (c) 0.8375 Ă— 10 MPa 5 IAS11. Ans.(b) E 2G (1 P ) or 1.25x10 = 2G(1+0.34) or G = 0.4664 Ă— 105 MPa
IAS12.
The Young's modulus of elasticity of a material is 2.5 times its modulus of rigidity. The Posson's ratio for the material will be: [IAS1997] (a) 0.25 (b) 0.33 (c) 0.50 (d) 0.75 E Â§ E Âˇ Â§ 2.5 Âˇ Â&#x; 1 P Â&#x;P Â¨ 1Â¸ Â¨ 1Â¸ 0.25 IAS12. Ans. (a) E 2G 1 P
2G 2G ÂŠ Âš ÂŠ 2 Âš
IAS13.
In a homogenous, isotropic elastic material, the modulus of elasticity E in terms of G and K is equal to [IAS1995, IES  1992] (a)
G 3K 9 KG
(b)
3G K 9 KG
(c)
9 KG G 3K
(d)
9 KG K 3G
IAS13. Ans. (c) The Elastic Constants E and K are related as ( P is the Poissonâ€™s ratio) [IAS1996] (a) E = 2k (1 â€“ 2 P ) (b) E = 3k (1 2 P ) (c) E = 3k (1 + P ) (d) E = 2K(1 + 2 P ) IAS14. Ans. (b) E = 2G (1 + P ) = 3k (1 2 P ) IAS14.
IAS15.
For an isotropic, homogeneous and linearly elastic material, which obeys Hooke's law, the number of independent elastic constant is: [IAS2000] (a) 1 (b) 2 (c) 3 (d) 6 IAS15. Ans. (b) E, G, K and Îź represent the elastic modulus, shear modulus, bulk modulus and poisons ratio respectively of a â€˜linearly elastic, isotropic and homogeneous material.â€™ To express the stress â€“ strain relations completely for this material; at least any two of the four must be known. IAS16.
E
2G 1 P 3K 1 3P
9 KG 3K G
The moduli of elasticity and rigidity of a material are 200 GPa and 80 GPa, respectively. What is the value of the Poisson's ratio of the material? [IAS2007] (a) 0Âˇ30 (b) 0Âˇ26 (c) 0Âˇ25 (d) 0Âˇ24
IAS16. Ans. (c) E = 2G (1+ P ) or
P=
E 1 2G
200 1 0.25 2 u 80
Stresses in compound strut IAS17.
The reactions at the rigid supports at A and B for the bar loaded as shown in the figure are respectively. [IES2002; IAS2003] (a) 20/3 kN,10/3 Kn (b) 10/3 kN, 20/3 kN (c) 5 kN, 5 kN (d) 6 kN, 4 kN
Page 38 of 429
Chapter1
Stress and Strain
S K Mondal’s
IAS17. Ans. (a) Elongation in AC = length reduction in CB
RA u 1 AE
RB u 2 AE
And RA + RB = 10
Thermal effect IAS18.
A steel rod 10 mm in diameter and 1m long is heated from 20°C to 120°C, E = 200 GPa and ǂ = 12 × 106 per °C. If the rod is not free to expand, the thermal stress developed is: [IAS2003, IES1997, 2000, 2006] (a) 120 MPa (tensile) (b) 240 MPa (tensile) (c) 120 MPa (compressive) (d) 240 MPa (compressive)
12 u 10 u 200 u 10 u 120 20
6
IAS18. Ans. (d) D E't
3
240MPa
It will be compressive as elongation restricted. IAS19.
A. steel rod of diameter 1 cm and 1 m long is heated from 20°C to 120°C. Its
D 12 u106 / K and
E=200 GN/m2. If the rod is free to expand, the thermal stress developed in it is: [IAS2002] (a) 12 × 104 N/m2 (b) 240 kN/m2 (c) zero (d) infinity IAS19. Ans. (c) Thermal stress will develop only if expansion is restricted. IAS20.
Which one of the following pairs is NOT correctly matched? [IAS1999] (E = Young's modulus, ǂ = Coefficient of linear expansion, T = Temperature rise, A = Area of crosssection, l= Original length) ….. ( DTl G ) (a) Temperature strain with permitted expansion G DTE (b) Temperature stress ….. DTEA (c) Temperature thrust ….. (d) Temperature stress with permitted expansion
…..
E (D Tl G ) l
IAS20. Ans. (a) Dimensional analysis gives (a) is wrong
Impact loading IAS21.
Match List I with List II and select the correct answer using the codes given below the lists: [IAS1995] List I (Property) List II (Testing Machine) A. Tensile strength 1. Rotating Bending Machine B. Impact strength 2. ThreePoint Loading Machine C. Bending strength 3. Universal Testing Machine D. Fatigue strength 4. Izod Testing Machine
Codes: (a) (c) IAS21. Ans. (d)
A 4 2
B 3 1
C 2 4
D 1 3
(b) (d)
Page 39 of 429
A 3 3
B 2 4
C 1 2
D 4 1
Chapter1
Stress and Strain
S K Mondalâ€™s
Tensile Test IAS22.
A mild steel specimen is tested in tension up to fracture in a Universal Testing Machine. Which of the following mechanical properties of the material can be evaluated from such a test? [IAS2007] 1. Modulus of elasticity 2. Yield stress 3. Ductility 4. Tensile strength 5. Modulus of rigidity Select the correct answer using the code given below: (a) 1, 3, 5 and 6 (b) 2, 3, 4 and 6 (c) 1, 2, 5 and 6 (d) 1, 2, 3 and 4 IAS22. Ans. (d) IAS23.
In a simple tension test, Hooke's law is valid upto the (a) Elastic limit (b) Limit of proportionality (c) Ultimate stress IAS23. Ans. (b)
[IAS1998] (d) Breaking point
IAS24.
Lueder' lines on steel specimen under simple tension test is a direct indication of yielding of material due to slip along the plane [IAS1997] (a) Of maximum principal stress (b) Off maximum shear (c) Of loading (d) Perpendicular to the direction of loading IAS24. Ans. (b) IAS25.
The percentage elongation of a material as obtained from static tension test depends upon the [IAS1998] (a) Diameter of the test specimen (b) Gauge length of the specimen (c) Nature of endgrips of the testing machine (d) Geometry of the test specimen IAS25. Ans. (b) IAS26.
Match ListI (Types of Tests and Materials) with ListII (Types of Fractures) and select the correct answer using the codes given below the lists: List I ListII [IES2002; IAS2004] (Types of Tests and Materials) (Types of Fractures) A. Tensile test on CI 1. Plain fracture on a transverse plane B. Torsion test on MS 2. Granular helecoidal fracture C. Tensile test on MS 3. Plain granular at 45Â° to the axis D. Torsion test on CI 4. Cup and Cone 5. Granular fracture on a transverse plane Codes: A B C D A B C D (a) 4 2 3 1 (c) 4 1 3 2 (b) 5 1 4 2 (d) 5 2 4 1 IAS26. Ans. (d) IAS27.
Assertion (A): For a ductile material stressstrain curve is a straight line up to the yield point. [IAS2003] Reason (R): The material follows Hooke's law up to the point of proportionality. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS27. Ans. (d) IAS28.
Assertion (A): Stressstrain curves for brittle material do not exhibit yield point. [IAS1996] Reason (R): Brittle materials fail without yielding. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS28. Ans. (a) Up to elastic limit. Page 40 of 429
Chapter1 IAS29.
Stress and Strain
Match List I (Materials) with List II (StressStrain curves) and select the correct answer using the codes given below the Lists: [IAS2001]
Codes: (a) (c) IAS29. Ans. (b) IAS30.
S K Mondalâ€™s
A 3 2
B 1 4
C 4 3
D 1 1
(b) (d)
A 3 4
B 2 1
C 4 3
D 2 2
The stressstrain curve of an ideal elastic strain hardening material will be as
[IAS1998] IAS30. Ans. (d)
IAS31.
An idealised stressstrain curve for a perfectly plastic material is given by
[IAS1996] IAS31. Ans. (a) IAS32.
Match List I with List II and select the correct answer using the codes given below the Lists: [IAS2002] List I List Page 41 II of 429 A. Ultimate strength 1. Internal structure
Chapter1
Stress and Strain
B. Natural strain C. Conventional strain D. Stress Codes: A B (a) 1 2 (c) 1 3 IAS32. Ans. (a)
C 3 2
S K Mondalâ€™s
2. Change of length per unit instantaneous length 3. Change of length per unit gauge length 4. Load per unit area D A B C D 4 (b) 4 3 2 1 4 (d) 4 2 3 1
IAS33.
What is the cause of failure of a short MS strut under an axial load? [IAS2007] (a) Fracture stress (b) Shear stress (c) Buckling (d) Yielding IAS33. Ans. (d) In compression tests of ductile materials fractures is seldom obtained. Compression is accompanied by lateral expansion and a compressed cylinder ultimately assumes the shape of a flat disc. IAS34.
Match List I with List II and select the correct answer using the codes given the lists: [IAS1995] List I List II A. RigidPerfectly plastic
B.
ElasticPerfectly plastic
C.
RigidStrain hardening
D.
Linearly elastic
Codes: (a) (c) IAS34. Ans. (a)
A 3 3
B 1 1
C 4 2
D 2 4
(b) (d)
A 1 1
B 3 3
C 2 4
D 4 2
IAS35.
Which one of the following materials is highly elastic? [IAS1995] (a) Rubber (b) Brass (c) Steel (d) Glass IAS35. Ans. (c) Steel is the highly elastic material because it is deformed least on loading, and regains its original from on removal of the load.
IAS36. Assertion (A): Hooke's law is the constitutive law for a linear elastic material. Reason (R) Formulation of the theory of elasticity requires the hypothesis that there exists a unique unstressed state of the body, to which the body returns whenever all the forces are removed. [IAS2002] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS36. Ans. (a) IAS37.
Page 42 of 429 Consider the following statements:
[IAS2002]
Chapter1
Stress and Strain
S K Mondal’s
1. There are only two independent elastic constants. 2. Elastic constants are different in orthogonal directions. 3. Material properties are same everywhere. 4. Elastic constants are same in all loading directions. 5. The material has ability to withstand shock loading. Which of the above statements are true for a linearly elastic, homogeneous and isotropic material? (a) 1, 3, 4 and 5 (b) 2, 3 and 4 (c) 1, 3 and 4 (d) 2 and 5 IAS37. Ans. (a) IAS38.
Which one of the following pairs is NOT correctly matched? [IAS1999] (a) Uniformly distributed stress …. Force passed through the centroid of the crosssection (b) Elastic deformation …. Work done by external forces during deformation is dissipated fully as heat (c) Potential energy of strain …. Body is in a state of elastic deformation (d) Hooke's law …. Relation between stress and strain IAS38. Ans. (b) A tensile bar is stressed to 250 N/mm2 which is beyond its elastic limit. At this stage the strain produced in the bar is observed to be 0.0014. If the modulus of elasticity of the material of the bar is 205000 N/mm2 then the elastic component of the strain is very close to [IAS1997] (a) 0.0004 (b) 0.0002 (c) 0.0001 (d) 0.00005 IAS39. Ans. (b) IAS39.
Page 43 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
Previous Conventional Questions with Answers Conventional Question IES2010 Q.
If a load of 60 kN is applied to a rigid bar suspended by 3 wires as shown in the above figure what force will be resisted by each wire? The outside wires are of Al, crosssectional area 300 mm2 and length 4 m. The central wire is steel with area 200 mm2 and length 8 m. Initially there is no slack in the wires E
2 u 105 N / mm 2 for Steel
0.667 u 105 N / mm 2 for Aluminum
[2 Marks]
Ans.
Aluminium wire FA1
FA1
FSt
Steel wire
60kN
P
a A1 ast EA1
Est
60 kN
300mm 2 l A1 200mm 2 lst
4m 8m
0.667 u 105 N / mm 2
2 u 105 N / mm 2
Force balance along vertical direction 2FA1 Fst 60 kN (1) Elongation will be same in all wires because rod is rigid remain horizontal loading
FA1 u l A1 a Al .EAl
Fst .lst ast .Est
(2)
Fst u 8
FA1 u 4 5
300 u 0.667 u 10 FA1 1.0005 Fst
200 u 2 u 105 (3)
Page 44 of 429
after
Chapter1
Stress and Strain
FA1
60 u 103 3.001
Fst
From equation (1)
S K Mondalâ€™s
19.99 kN or 20 kN
20 kN
FA1 Fst
20 kN Â˝ Âž Answer. 20 kN Âż
Conventional Question GATE The diameters of the brass and steel segments of the axially loaded bar shown in figure are 30 mm and 12 mm respectively. The diameter of the hollow section of the brass segment is 20 mm. Determine: (i) The maximum normal stress in the steel and brass (ii) The displacement of the free end ; Take Es = 210 GN/m2 and Eb = 105 GN/m2 Question:
Answer:
As
S 4
u 12
A b BC
S
Ab CD
S
4 4
2
36S mm2
u 30
2
36S u 10 6 m2
225S mm2
u 302 202
225S u 106 m2
125S mm2
125S u 106 m2
(i) The maximum normal stress in steel and brass:
10 u 103 u 10 6 MN / m2 88.42MN / m2 6 36S u 10 5 u 103 u 10 6 MN / m2 7.07MN / m2 V b BC 225S u 10 6 5 u 103 u 106 MN / m2 12.73MN / m2 V b CD 125S u 10 6
Vs
(ii) The displacement of the free end:
Gl
G ls AB G lb BC G lb CD 88.42 u 0.15 7.07 u 0.2 12.73 u 0.125 9 9 6 6 210 u 10 u 10 105 u 10 u 10 105 u 109 u 10 6 9.178 u 105 m
Â§ Â¨ G l ÂŠ
Vl Âˇ E Â¸Âš
0.09178 mm
Conventional Question IES1999 Question: Answer:
Distinguish between fatigue strength and fatigue limit. Fatigue strength as the value of cyclic stress at which failure occurs after N cycles. And fatigue limit as the limiting value of stress at which failure occurs as N becomes very large (sometimes called infinite cycle)
Conventional Question IES1999
Page 45 of 429
Chapter1 Question: Answer:
Stress and Strain
S K Mondalâ€™s
List at least two factors that promote transition from ductile to brittle fracture. (i) With the grooved specimens only a small reduction in area took place, and the appearance of the facture was like that of brittle materials. (ii) By internal cavities, thermal stresses and residual stresses may combine with the effect of the stress concentration at the cavity to produce a crack. The resulting fracture will have the characteristics of a brittle failure without appreciable plastic flow, although the material may prove ductile in the usual tensile tests.
Conventional Question IES1999 Question: Answer:
Distinguish between creep and fatigue. Fatigue is a phenomenon associated with variable loading or more precisely to cyclic stressing or straining of a material, metallic, components subjected to variable loading get fatigue, which leads to their premature failure under specific conditions. When a member is subjected to a constant load over a long period of time it undergoes a slow permanent deformation and this is termed as ''Creep''. This is dependent on temperature.
Conventional Question IES2008 Question: Answer:
What different stresses setup in a bolt due to initial tightening, while used as a fastener? Name all the stresses in detail. (i) When the nut is initially tightened there will be some elongation in the bolt so tensile stress will develop. (ii) While it is tightening a torque across some shear stress. But when tightening will be completed there should be no shear stress.
Conventional Question IES2008 Question:
A Copper rod 6 cm in diameter is placed within a steel tube, 8 cm external diameter and 6 cm internal diameter, of exactly the same length. The two pieces are rigidly fixed together by two transverse pins 20 mm in diameter, one at each end passing through both rod and the tube. Calculated the stresses induced in the copper rod, steel tube and the pins if the temperature of the combination is raised by 50oC. [Take ES=210 GPa,
Bs 0.0000115 / o C ; Ec=105 GPa, Bc 0.000017 / o C ]
Answer:
Tc Ts
%t (Bc Bs ) Ec Es 2
Qd 2 Q Â?Âž 6 ÂŹÂ 2 3 2 = ÂžÂž ÂÂ m 2.8274 q10 m 4 4 Â&#x;100 ÂŽ 2 2 Qd 2 Q Â ÂĄÂ?Âž 8 ÂŹÂ Â?Âž 6 ÂŹÂ ÂŻÂ° 2 3 2 Area of steel tube (A s ) = ÂĄÂžÂž Â Âž Â m 2.1991q10 m 4 4 ÂĄÂ˘Â&#x;100 ÂŽÂ Â&#x;Âž100 ÂŽÂ Â°ÂąÂ° Rise in temperature,%t 50o C Area of copper rod(A c )
Free expansion of copper bar=Bc L+t Free expansion of steel tube =Page +tof 429 Bs L46
Chapter1
Stress and Strain
S K Mondalâ€™s
Difference in free expansion = Bc Bs L+t
= 1711.5 Ă—106 q L q 50=2.75Ă—104L m A compressive force (P) exerted by the steel tube on the copper rod opposed the extra expansion of the copper rod and the copper rod exerts an equal tensile force P to pull the steel tube. In this combined effect reduction in copper rod and increase in length of steel tube equalize the difference in free expansions of the combined system. Reduction in the length of copper rod due to force P Newton=
+L C
PL PL m Ac Ec 2.8275 q103 105 q109
Increase in length of steel tube due to force P PL P .L m +L S As Es 2.1991 q103 210 q109
Difference in length is equated
+L c +L s 2.75 q104 L PL P.L
2.75 q104 L 3 3 9 9 q q q q 2.8275 10 105 10 2.1991 10 210 10
Or P = 49.695 kN
49695 P MPa=17.58MPa 2.8275 q103 Ac P 49695 Stress in steel tube, Ts MPa 22.6MPa As 2.1991q103 Stress in copper rod, Tc
Since each of the pin is in double shear, shear stress in pins ( U pin ) =
P 49695 = 79 MPa 2 q Apin 2 q Q 0.02 2 4
Conventional Question IES2002 Question: Answer:
Why are the bolts, subjected to impact, made longer? If we increase length its volume will increase so shock absorbing capacity will increased.
Conventional Question IES2007 Question:
Answer:
Explain the following in brief: (i) Effect of size on the tensile strength (ii) Effect of surface finish on endurance limit. (i) When size of the specimen increases tensile strength decrease. It is due to the reason that if size increases there should be more change of defects (voids) into the material which reduces the strength appreciably. (ii) If the surface finish is poor, the endurance strength is reduced because of scratches present in the specimen. From the scratch crack propagation will start.
Conventional Question IES2004 Question:
Mention the relationship between three elastic constants i.e. elastic modulus (E), rigidity modulus (G), and bulk modulus (K) for any Elastic material. How is the Poisson's ratio ( Č? ) related to these modulli? Answer: E
9KG 3K G
9KG 3K + G Page 47 of 429
E 3K (1 2Îź) = 2G(1 + Îź) =
Chapter1
Stress and Strain
S K Mondalâ€™s
Conventional Question IES1996 Question:
The elastic and shear moduli of an elastic material are 2Ă—1011 Pa and 8Ă—1010 Pa respectively. Determine Poisson's ratio of the material.
Answer:
We know that E = 2G(1+ Îź ) = 3K(1  2Îź) =
or,1 Îź or Îź
9KG 3K + G
E 2G
E 2 q1011 1 1 0.25 2G 2 q (8 q1010 )
Conventional Question IES2003 Question:
A steel bolt of diameter 10 mm passes through a brass tube of internal diameter 15 mm and external diameter 25 mm. The bolt is tightened by a nut so that the length of tube is reduced by 1.5 mm. If the temperature of the assembly is raised by 40oC, estimate the axial stresses the bolt and the tube before and after heating. Material properties for steel and brass are:
ES 2q105 N / mm 2
BS 1.2q105 / o C and Eb= 1Ă—105 N/mm2 B b=1.9Ă—1015/oC
Answer:
ĘŒ Area of steel bolt (A s )= q (0.010)2 m 2 7.854 q105 m 2 4 ĘŒ Area of brass tube (A b )= Â ÂĄÂ˘(0.025)2 (0.015)2 ÂŻÂ°Âą 3.1416 q104 4 Stress due to tightening of the nut Compressive force on brass tube= tensile fore on steel bolt or, Äąb Ab ÄąS As
or , Eb.
Â ÂŻ ÂĄ Â° ÂĄ Äą Äą Â° ÂĄ' E= Â° ÂĄ Â‰ Â?Âž +L ÂŹÂ Â° ÂĄ ÂžÂž ÂÂ Â°Â° ÂĄÂ˘ Â&#x; L ÂŽÂą
(%l )b .Ab Äąs As A
Let assume total length (A )=1m
(1.5 q103 ) q 3.1416 q104 Ts q 7.854Ă—105 1 Ts 600 MPa (tensile )
Therefore (1Ă—105 q106 ) q or
and Tb =Eb.
(%l )b (1.5 q103 ) MPa 150MPa (Compressive ) (1Ă—105 ) q 1 A Page 48 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
So before heating Stress in brass tube (Äąb ) 150MPa (compressive ) Stress in steel bolt(Äą s ) 600MPa (tensile) Stress due to rise of temperature Let stress
Äą 'b & Äą 's
are due to brass tube and steel bolt.
If the two members had been free to expand, Free expansion of steel = Bs q+t q1 Free expansion of brass tube = Since
Bb q+t q1
Bb Äąs free expansion of copper is greater than the free expansion of steel. But
they are rigidly fixed so final expansion of each members will be same. Let us assume this final expansion is ' ÄŻ ', The free expansion of brass tube is grater than ÄŻ , while the free expansion of steel is less than E . Hence the steel rod will be subjected to a tensile stress while the brass tube will be subjected to a compressive stress. For the equilibrium of the whole system, Total tension (Pull) in steel =Total compression (Push) in brass tube.
As 7.854 q105 ' ÄąS 0.25ÄąS' Ab 3.14 q104 Final expansion of steel =final expansion of brass tube Äą 'b Ab Äąs' As or , Äąb' Äą's q
Bs (+t ).1
T' Äą s' q1 Bb (+t ) q1 b q1 Es Eb
or , 1.2 q105 q 40 q1
Äąs' Äą 'b 5 q q q (ii ) (1.9 10 ) 40 1 2 q105 q106 1q105 q106
From(i) & (ii) we get Â 1 0.25 ÂŻ
11 Â° 2.8 q104 Äąs' ÂĄ 11 ÂĄ 2 q10 10 Â°Âą Â˘ or , Äą s' 37.33 MPa (Tensile stress) or, Äą b' = 9.33MPa (compressive) Therefore, the final stresses due to tightening and temperature rise
Stress in brass tube =Äą b +Äąb' =150+9.33MPa=159.33MPa Stress in steel bolt =Äą s +Äą 's = 600 + 37.33 = 637.33MPa. Conventional Question IES1997 Question:
Answer:
A Solid right cone of axial length h is made of a material having density S and elasticity modulus E. It is suspended from its circular base. Determine its elongation due to its self weight. See in the figure MNH is a solid right cone of length 'h' . Let us assume its wider end of diameterâ€™dâ€™ fixed rigidly at MN. Now consider a small strip of thickness dy at a distance y from the lower end. Let 'ds' is the diameter of the strip.
1 Â? ĘŒd 2 ÂŹ = Weight of portion UVH= ÂžÂžÂž s ÂÂÂ y q Sg (i ) 3 ÂžÂ&#x; 4 ÂŽÂ Page 49 of 429
Chapter1
Stress and Strain
S K Mondalâ€™s
From the similar triangles MNH and UVH, A MN d y UV ds or , ds
d .y ( ii ) A
= Stress at section UV =
force at UV Weight of UVH Â? ĘŒd s2 ÂŹÂ cross sec tion area at UV ÂžÂž Â ÂžÂ&#x; 4 ÂŽÂÂ
1 ĘŒd s2 . .y .Sg 1 3 4 y Sg = Â? ĘŒd 2 ÂŹÂ 3 ÂžÂž s Â ÂžÂ&#x; 4 ÂŽÂÂ Â?1 ÂŹ ÂžÂž y Sg ÂÂ.dy ÂžÂ&#x; 3 ÂŽÂ So, extension in dy= E h
= Total extension of the bar =Â¨ 0
1 y Sgdy Sgh 2 3 E 6E
From stressstrain relation ship E=
ÄŻ ÄŻ E.A or , d A d A E Â‰ A
Conventional Question IES2004 Question:
Answer:
Which one of the three shafts listed hare has the highest ultimate tensile strength? Which is the approximate carbon content in each steel? (i) Mild Steel (ii) cast iron (iii) spring steel Among three steel given, spring steel has the highest ultimate tensile strength. Approximate carbon content in (i) Mild steel is (0.3% to 0.8%) (ii) Cost iron (2% to 4%) (iii) Spring steel (0.4% to 1.1%)
Conventional Question IES2003 Question: Answer:
If a rod of brittle material is subjected to pure torsion, show with help of a sketch, the plane along which it will fail and state the reason for its failure. Brittle materials fail in tension. In a torsion test the maximum tensile test Occurs at 45Â° to the axis of the shaft. So failure will occurs along a 45o to the axis of the shaft. So failure will occurs along a 45Â° helix
X
X 50 of 429 So failures will occurs according toPage 45Â° plane.
Chapter1
Stress and Strain
S K Mondalâ€™s
Conventional Question IAS1995 Question:
Answer:
The steel bolt shown in Figure has a thread pitch of 1.6 mm. If the nut is initially tightened up by hand so as to cause no stress in the copper spacing tube, calculate the stresses induced in the tube and in the bolt if a spanner is then used to turn the nut through 90Â°.Take Ec and Es as 100 GPa and 209 GPa respectively. Given: p = 1.6 mm, Ec= 100 GPa ; Es = 209 CPa.
Stresses induced in the tube and the bolt, V c ,V s :
As As
2
S
Â§ 10 Âˇ uÂ¨ 7.584 u 10 5 m2 Â¸ 4 ÂŠ 1000 Âš 2 2 S ÂŞÂ§ 18 Âˇ Â§ 12 Âˇ Âş 5 2 u ÂŤÂ¨ Âť 14.14 u 10 m Â¸ Â¨ Â¸ 4 ÂŤÂŹÂŠ 1000 Âš ÂŠ 1000 Âš ÂťÂź
Tensile force on steel bolt, Ps = compressive force in copper tube, Pc = P Also, Increase in length of bolt + decrease in length of tube = axial displacement of nut
i,e or or or
G l s G l c
1.6 u
Pl Pl A sEs A cEc
90 360
0.4mm
0.4 u 103
0.4 u 10 3 m
ls
lc
l
1 1 Â§ 100 Âˇ ÂŞ Âş PuÂ¨ Â¸ 5 5 9 9 Âť ÂŤ 14.14 u 10 u 100 u 10 Âź ÂŠ 1000 Âš ÂŹ 7.854 u 10 u 209 u 10 P 30386N P P 386.88MPa and 214.89MPa ? As Ac
0.4 u 103
Conventional Question AMIE1997 Question:
Answer:
A steel wire 2 m long and 3 mm in diameter is extended by 0Âˇ75 mm when a weight W is suspended from the wire. If the same weight is suspended from a brass wire, 2Âˇ5 m long and 2 mm in diameter, it is elongated by 4 64 mm. Determine the modulus of elasticity of brass if that of steel be 2.0 Ă— 105 N / mm2 Given, ls 2 m, ds = 3 mm, G ls 0Âˇ75 mm; Es = 2Âˇ0 Ă— 105 N / mm2; lb 2.5 m, db =2 mm G lb
4.64m m and let modulus of elasticity of brass = Eb
Hooke's law gives, G l
Pl AE
[Symbol has usual meaning]
Case I: For steel wire:
Page 51 of 429
Chapter1
Stress and Strain
G ls
S K Mondal’s
Pls A sEs P u 2 u 1000
or 0.75
 (i)
1 §S 2· 5 ¨ 4 u 3 ¸ u 2.0 u 10 u 2000 © ¹
Case II: For bass wire:
G lb 4.64
or
Plb A bEb P u 2.5 u 1000
 (ii)
§S 2· ¨ 4 u 2 ¸ u Eb © ¹ P
1 §S · 4.64 u ¨ u 22 ¸ u Eb u 2500 ©4 ¹
From (i) and (ii), we get
1 §S · 0.75 u ¨ u 32 ¸ u 2.0 u 105 u 2000 ©4 ¹ 5 or Eb 0.909 u 10 N / mm2
1 §S · 4.64 u ¨ u 22 ¸ u Eb u 2500 ©4 ¹
Conventional Question AMIE1997 Question:
A steel bolt and sleeve assembly is shown in figure below. The nut is tightened up on the tube through the rigid end blocks until the tensile force in the bolt is 40 kN. If an external load 30 kN is then applied to the end blocks, tending to pull them apart, estimate the resulting force in the bolt and sleeve.
2
Answer:
§ 25 · 4.908 u 104 m2 ¨ 1000 ¸ © ¹ 2 2 S ª§ 62.5 · § 50 · º Area of steel sleeve, A s «¨ » 4 ¬«© 1000 ¸¹ ¨© 1000 ¸¹ ¼» Area of steel bolt, A b
1.104 u 10 3 m2
Forces in the bolt and sleeve: (i) Stresses due to tightening the nut: Let V b = stress developed in steel bolt due to tightening the nut; and
V s = stress developed in steel sleeve due to tightening the nut. Tensile force in the steel bolt = 40 kN = 0·04 MN
Page 52 of 429
Chapter1
Stress and Strain
V b u Ab
S K Mondalâ€™s
0.04
or
V b u 4.908 u 104
?
Vb
0.04
0.04 4.908 u 104
81.5MN / m2 tensile
Compressive force in steel sleeve = 0Âˇ04 MN
V s u As
0.04
or
V s u 1.104 u 103
?
Vs
0.04
0.04 1.104 u 10 3
36.23MN / m2 compressive
(ii) Stresses due to tensile force: Let the stresses developed due to tensile force of 30 kN = 0Âˇ03 MN in steel bolt and sleeve be V 'b and V 's respectively. Then, V 'b u A b V 's u A s
0.03
4
V 'b u 4.908 u 10 V 's u 1.104 u 103
(i)
0.03
In a compound system with an external tensile load, elongation caused in each will be the same.
V 'b
G lb
Eb
V 'b
or G lb
Eb
V 's
and G ls
Es
But G lb ?
V 'b Eb
or
u lb u 0.5
Given,lb
500mm
0.5
u 0.4
Given,ls
400mm
0.4
Gs V 's
u 0.5
V 'b
Es
u 0.4
0.8V 's
Given,Eb
Es
(2)
Substituting this value in (1), we get
0.8V 's u 4.908 u 10 4 V 's u 1.104 u 10 3
V 's
gives and
V 'b
0.03
20MN / m tensile
2
0.8 u 20 16MN / m2 tensile
Re sulting stress in steel bolt,
V b r
V b V 'b
81.5 16
97.5MN / m2
Re sulting stress in steel sleeve,
V s r
V s V 's
36.23 20 16.23MN / m2 compressive
V b r u A b
Re sulting force in steel bolt,
97.5 u 4.908 u 104 Re sulting force in steel sleeve
0.0478MN tensile
V b r u A s 16.23 u 1.104 u 10 3
Page 53 of 429
0.0179MN compressive
2.
Principal Stress and Strain
Theory at a Glance (for IES, GATE, PSU) 2.1 States of stress
x
Uniaxial stress: only one nonzero principal stress, i.e. ǔ1
Right side figure represents Uniaxial state of stress.
x
Biaxial stress: one principal stress equals zero, two do not, i.e. ǔ1 > ǔ3 ; ǔ2 = 0
Right side figure represents Biaxial state of stress.
x
Triaxial
stress:
three
nonzero
principal stresses, i.e. ǔ1 > ǔ2 > ǔ3 Right side figure represents Triaxial state of stress.
x
Isotropic
stress:
three
principal
stresses are equal, i.e. ǔ1 = ǔ2 = ǔ3 Right side figure represents isotropic state of stress.
x
Axial stress: two of three principal stresses are equal, i.e. ǔ1 = ǔ2 or ǔ2 = ǔ3 Right side figure represents axial state of stress.
Page 54 of 429
Chapter2
x
Principal Stress and Strain
S K Mondal’s
Hydrostatic pressure: weight of column of fluid in interconnected pore spaces. Phydrostatic = ǒfluid gh (density, gravity, depth)
x
Hydrostatic stress: Hydrostatic stress is used to describe a state of tensile or compressive stress equal in all directions within or external to a body. Hydrostatic
Or
stress causes a change in volume of a material.
Shape
of
the
body
remains
unchanged i.e. no distortion occurs in the body. Right side figure represents Hydrostatic state of stress.
2.2 Uniaxial stress on oblique plane Let us consider a bar of uniform cross sectional area A under direct tensile load P giving rise to axial normal stress P/A acting on a cross section XX. Now consider another section given by the plane YY inclined at T with the XX. This is depicted in following three ways.
Fig. (a)
Fig. (b)
Fig. (c) Area of the YY Plane = a shear stress
A ; Let us assume the normal stress in the YY plane is cos T
Vn
and there is
W acting parallel to the YY plane.
Now resolve the force P in two perpendicular direction one normal to the plane YY = P cos T and another parallel to the plane YY = Pcosș Page 55 of 429
Chapter2
Principal Stress and Strain
Therefore equilibrium gives,
Vn
and
x
Wu
A cos T
P sin T
or
W
A cos T
P cos T
Note the variation of normal stress When T
Vn
P cos 2 T A
Vn
or
P sin T cos T A
S K Mondalâ€™s
W
or
P sin 2T 2A
and shear stress
0 , normal stress V n is maximum i.e. V n max
W with
P and shear stress W A
increased, the normal stress V n diminishes, until when T
T increased shear stress W increases to a maximum value W max diminishes to W
0 at T
the variation of T .
0, V n
0.
P at T 2A
S 4
0 . As T is But if angle
45o and then
90o
x
The shear stress will be maximum when sin 2T
x
And the maximum shear stress, W max
x
In ductile material failure in tension is initiated by shear stress i.e. the failure occurs across
1 or T
45o
P 2A
the shear planes at 45o (where it is maximum) to the applied load. Let us clear a concept about a common mistake: The angle T is not between the applied load and the plane. It is between the planes XX and YY. But if in any question the angle between the applied load and the plane is given donâ€™t take it as T . The angle between the applied load and the plane
Vn
is
90

T.
P cos2 (90 T ) and W A
In
this
case
you
have
to
use
the
above
formula
as
P sin(180 2T ) where T is the angle between the applied load and the 2A
plane. Carefully observe the following two figures it will be clear.
Let us take an example: A metal block of 100 mm2 cross sectional area carries an axial tensile load of 10 kN. For a plane inclined at 300 with the direction of applied load, calculate: (a) Normal stress (b) Shear stress (c) Maximum shear stress. Answer: Here T
90o 30o
60o Page 56 of 429
Chapter2
Principal Stress and Strain P 10 u 103 N cos2 T u cos2 60o 25MPa 2 100 mm A
(a) Normal stress V n
10 u 103 N u sin120o 2 2 u 100 mm
P sin2T 2A
(b) Shear stress W
x
10 u 103 N 2 u 100 mm 2
P 2A
(c) Maximum shear stress W max
S K Mondalâ€™s
43.3MPa
50MPa
Complementary stresses
Now if we consider the stresses on an oblique plane Yâ€™Yâ€™ which is perpendicular to the previous plane YY. The stresses on this plane are known as complementary stresses. Complementary normal stress is
V nc
and complementary shear stress is
the four stresses. To obtain the stresses previous equation. The angle
T 900
V nc
and
W c . The following figure shows all
W c we need only to replace T
by
T 900 in the
is known as aspect angle.
Therefore
P cos 2 90o T
A
V nc Wc
It is clear
P sin 2 90o T
2A
V nc V n
P A
and
P 2 sin T A
P sin 2T 2A
W c W
i.e. Complementary shear stresses are always equal in magnitude but opposite in sign.
x
Sign of Shear stress
For sign of shear stress following rule have to be followed: Page 57 of 429
Chapter2 The shear stress
W on
Principal Stress and Strain
S K Mondalâ€™s
any face of the element will be considered positive when it has a
clockwise moment with respect to a centre inside the element. If the moment is counterclockwise with respect to a centre inside the element, the shear stress in negative. Note: The convention is opposite to that of moment of force. Shear stress tending to turn clockwise is positive and tending to turn counter clockwise is negative.
Let us take an example: A prismatic bar of 500 mm2 cross sectional area is axially loaded with a tensile force of 50 kN. Determine all the stresses acting on an element which makes 300 inclination with the vertical plane. Answer: Take an small element ABCD in 300 plane as shown in figure below, Given, Area of crosssection, A = 500 mm2, Tensile force (P) = 50 kN
Normal stress on 30Â° inclined plane,
Shear stress on 30Â° planes, W
P 50Ă—103 N 2 Äą = cos Č™ = Ă—cos2 30o =75MPa (+ive means tensile). n
2 A 500 mm 50 u 103 N u sin 2 u 30o 43.3MPa 2 u 500 mm 2
P sin2T 2A
(+ive means clockwise) Complementary stress on T
90 30 120o
Normal stress on 1200 inclined plane, V nc
P cos2 T A
50 u 103 N u cos2 120o 25MPa 500 mm 2 (+ ive means tensile)
Shear stress on 1200 nclined plane, W c
P sin2T 2A
50 u 103 N u sin 2 u 120o
2 2 u 500 mm
43.3MPa
( ive means counter clockwise) State of stress on the element ABCD is given below (magnifying)
Page 58 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
2.3 Complex Stresses (2D Stress system) i.e. Material subjected to combined direct and shear stress We now consider a complex stress system below. The given figure ABCD shows on small element of material
Stresses in three dimensional element
Vx
and
Vy
Stresses in crosssection of the element
are normal stresses and may be tensile or compressive. We know that normal stress
may come from direct force or bending moment. comes from direct shear force or torsion and
W xy
W xy
is shear stress. We know that shear stress may
and
W yx
are complementary and
W xy = W yx Let
Vn
is the normal stress and
W
is the shear stress on a plane at angle T .
Considering the equilibrium of the element we can easily get
Normal stress V n
V x V y 2
V x V y 2
cos 2T W xy sin 2T
and
Shear stress W
Äąx Äąy 2
sin 2Č™  W xy cos 2Č™
Above two equations are coming from considering equilibrium. They do not depend on material Page 59 of 429
properties and are valid for elastic and in elastic behavior.
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
x Location of planes of maximum stress
V n max
(a) Normal stress,
Vn
For
wV n wT
maximum or minimum
or
V
x
Vy
2
(b) Shear stress, For
wW wT or
V
0, where V n
W
x
Vy
2
V
x
Vy
2
cos 2T W xy sin 2T
u sin 2T u 2 W xy cos 2T u 2
0
or tan2T p =
2W xy (V x V y )
W max
maximum or minimum
0, where W
Vx Vy 2
Vx Vy 2
sin 2T W xy cos 2T
cos 2T u 2 W xy sin 2T u 2
0
W xy
or cot 2T
Vx Vy
Let us take an example: At a point in a crank shaft the stresses on two mutually perpendicular planes are 30 MPa (tensile) and 15 MPa (tensile). The shear stress across these planes is 10 MPa. Find the normal and shear stress on a plane making an angle 300 with the plane of first stress. Find also magnitude and direction of resultant stress on the plane.
25MPa tensile , V y
Answer: Given V x
Therefore, Normal stress V n
Shear stress W
V x V y
15MPa tensile , W xy
V x V y
10MPa and 400
cos 2T W xy sin2T 2 2 30 15 30 15 cos 2 u 30o 10 sin 2 u 30o 34.91 MPa 2 2
V x V y
sin2T W xy cos 2T 2 30 15 sin 2 u 30o 10cos 2 u 30o 1.5MPa 2
Resultant stress V r
34.91
and Obliquity I , tanI
W Vn
2
1.52
1.5 Â&#x;I 34.91
34.94MPa 2.460
Page 60 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
2.4 Biaxial stress Let us now consider a stressed element ABCD where W xy =0, i.e. only V x and V y is there. This type of stress is known as biaxial stress. In the previous equation if you put W xy =0 we get Normal stress,
V n and shear stress, W
on a plane at angle T .
Vx V y
Vn
V x V y
cos 2T
x
Normal stress ,
x
Shear/Tangential stress,
x
For complementary stress, aspect angle =
x
Aspect angle â€˜ Č™ â€™ varies from 0 to
x
Normal stress
V x (T
2
W
2
V x V y 2
sin 2T
T 900
S /2
V n varies between the values
0) & V y (T
S / 2)
Let us take an example: The principal tensile stresses at a point across two perpendicular planes are 100 MPa and 50 MPa. Find the normal and tangential stresses and the resultant stress and its obliquity on a plane at 200 with the major principal plane Answer: Given V x
Normal stress, V n
Shear stress, W
100MPa tensile , V y
Vx Vy 2
V x V y
Resultant stress V r
2
V x V y
sin2T
942 162
50MPa tensile and T
200
100 50 100 50 cos 2 u 20o
2 2
cos 2T
2 100 50 sin 2 u 200 2
16MPa
95.4MPa
Â§W Âˇ Â§ 16 Âˇ Therefore angle of obliquity, I tan1 Â¨ Â¸ tan1 Â¨ Â¸ 9.70 ÂŠ 94 Âš ÂŠ Vn Âš
Page 61 of 429
94MPa
Chapter2
x
Principal Stress and Strain
S K Mondalâ€™s
We may derive uniaxial stress on oblique plane from
Vx V y
Vn and
2 Äąx Äą y
W
2
Just put V y
V x V y 2
cos 2T W xy sin 2T
sin 2Č™  W xy cos 2Č™
0 and W xy =0
Therefore,
Vn
Vx 0 2
and W
Vx 0
Vx 0 2
2
cos 2T
sin 2T
Vx 2
1 V x 1 cos 2T V x cos2 T 2
sin 2T
2.5 Pure Shear
x
Pure shear is a particular case of biaxial stress where
Vx
V y
Note: V x or V y which one is compressive that is immaterial but one should be tensile and other
should
be
compressive
and
equal
magnitude.
V x 100MPa then
If
V y must be 100MPa otherwise if V y 100MPa then V x must be 100MPa .
x
In case of pure shear on 45o planes
W max x
rV x
; Vn
0 and V nc
0
We may depict the pure shear in an element by following two ways Page 62 of 429
Chapter2
Principal Stress and Strain
S K Mondal’s
(a) In a torsion member, as shown below, an element ABCD is in pure shear (only shear stress is present in this element) in this member at 45o plane an element AcBcC cDc is also in pure shear where V x
V y but in this element no shear stress is there.
(b) In a biaxial state of stress a member, as shown below, an element ABCD in pure shear where V x
V y but in this element no shear stress is there and an element AcBcC cDc at
45o plane is also in pure shear (only shear stress is present in this element).
Let us take an example: See the in the Conventional question answer section in this chapter and the question is “Conventional Question IES2007”
2.6 Stress Tensor
x
State of stress at a point ( 3D)
Stress acts on every surface that passes through the point. We can use three mutually perpendicular planes to describe the stress state at the point, which we approximate as a cube each of the three planes has one normal component & two shear components therefore, 9 components necessary to define stress at a point 3 normal and 6 shear stress. Therefore, we need nine components, to define the state of stress at a point
V x W xy W xz V y W yx W yz V z W zx W zy For cube to be in equilibrium (at rest: not moving, not spinning) Page 63 of 429
Chapter2
Principal Stress and Strain
S K Mondal’s
W xy
W yx
If they don’t offset, block spins therefore,
W xz W yz
W zx W zy
only six are independent.
The nine components (six of which are independent) can be written in matrix form
V ij
§ V xx ¨ ¨ V yx ¨V © zx
V xy V xz · ¸ V yy V yz ¸ or W ij V zy V zz ¸¹
§W xx W xy ¨ ¨W yx W yy ¨W © zx W zy
W xz · ¸ W yz ¸ W zz ¸¹
§ V x W xy ¨ ¨W yx V y ¨W © zx W zy
W xz · ¸ W yz ¸ V z ¸¹
§ V 11 V 12 V 13 · ¨ ¸ ¨ V 21 V 22 V 23 ¸ ¨V ¸ © 31 V 32 V 33 ¹
This is the stress tensor Components on diagonal are normal stresses; off are shear stresses
x
State of stress at an element (2D)
2.7 Principal stress and Principal plane x
When examining stress at a point, it is possible to choose three mutually perpendicular planes on which no shear stresses exist in three dimensions, one combination of orientations for the three mutually perpendicular planes will cause the shear stresses on all three planes to go to zero this is the state defined by the principal stresses. Page 64 of 429
Chapter2 x
Principal Stress and Strain
S K Mondalâ€™s
Principal stresses are normal stresses that are orthogonal to each other
x
Principal planes are the planes across which principal stresses act (faces of the cube) for principal stresses (shear stresses are zero)
x
Major Principal Stress
V1 x
2
Vx V y 2
Âˇ 2 W Â¸Â¸ xy Âš
Â§ V x V y Â¨ Â¨ 2 ÂŠ
2
Âˇ 2 W Â¸Â¸ xy Âš
Position of principal planes
tan2T p = x
Â§ V x V y Â¨ Â¨ 2 ÂŠ
Minor principal stress
V2 x
Vx V y
2
2W xy (V x V y )
Maximum shear stress
W max
V1 V 2 2
Â§ V x V y Â¨Â¨ ÂŠ 2
2
Âˇ 2 W Â¸Â¸ xy Âš
Let us take an example: In the wall of a cylinder the state of stress is given by,
Vx
85MPa compressive , V y
25MPa tensile and shear stress W xy 60MPa
Calculate the principal planes on which they act. Show it in a figure. Answer: Given V x
85MPa,V y
25MPa,W xy
60MPa
Page 65 of 429
Chapter2
Principal Stress and Strain
Major principal stress V 1
Vx Vy 2
S K Mondalâ€™s
2
Â§ V x V y Âˇ 2 Â¨ Â¸ W xy 2 ÂŠ Âš 2
85 25 Â§ 85 25 Âˇ 2 Â¨ Â¸ 60 2 2 ÂŠ Âš
Minor principal stress V 2
Vx Vy 2
51.4MPa
2
Â§ V x V y Âˇ 2 Â¨ Â¸ W xy 2 ÂŠ Âš 2
85 25 Â§ 85 25 Âˇ 2 Â¨ Â¸ 60 2 2 ÂŠ Âš 111.4 MPa i.e. 111.4 MPa Compressive
For principalplanes tan2TP or TP
2W xy
V x V y
2 u 60 85 25
240 it is for V 1
Complementary plane TP c T P 90 660 it is for V 2 The Figure showing state of stress and principal stresses is given below
The direction of one principle plane and the principle stresses acting on this would be V
1
when is
acting normal to this plane, now the direction of other principal plane would be 900 + T p because the principal planes are the two mutually perpendicular plane, hence rotate the another plane 900 + T p in the same direction to get the another plane, now complete the material element as T p is negative that means we are measuring the angles in the opposite direction to the reference plane BC. The following figure gives clear idea about negative and positive T p .
Page 66 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
2.8 Mohr's circle for plane stress x
The transformation equations of plane stress can be represented in a graphical form which is popularly known as Mohr's circle.
x
Though the transformation equations are sufficient to get the normal and shear stresses on any plane at a point, with Mohr's circle one can easily visualize their variation with respect to plane orientation Ç‰.
x
Equation of Mohr's circle
We know that normal stress,
And Tangential stress,
Ä˛
Â§
Rearranging we get, Â¨ V n Â¨
Vn
Äąx Äą y 2
2
V x V y 2
2
Äąx Äą y
Ä˛
2
cos 2T W xy sin 2T
sin2Č™  Ä˛ xy cos 2Č™
V x V y Âˇ V x V y
ÂŠ
and
Vx V y
Â¸Â¸ Âš
2
cos 2T W xy sin 2T â€Śâ€Śâ€Śâ€Śâ€Ś(i)
sin2Č™  Ä˛ xy cos 2Č™ â€Śâ€Śâ€Śâ€Śâ€Ś(ii)
A little consideration will show that the above two equations are the equations of a circle with
Vn
and U as its coordinates and 2Ç‰ as its parameter. If the parameter 2Ç‰ is eliminated from the equations, (i) & (ii) then the significance of them will become clear.
Vx V y
V avg Or
V
2 2
n
and R =
V avg W xy2
Â§ V x V y Â¨Â¨ ÂŠ 2
2
Âˇ 2 Â¸Â¸ W xy Âš
R2
It is the equation of a circle with centre,
Â§Vx V y
V avg ,0 i.e. Â¨Â¨ ÂŠ
Page 67 of 429
2
Âˇ ,0 Â¸ Â¸ Âš
Chapter2
Principal Stress and Strain
and radius,
R
Â§ V x V y Â¨Â¨ 2 ÂŠ
S K Mondalâ€™s
2
Âˇ 2 Â¸Â¸ W xy Âš
x Construction of Mohrâ€™s circle Convention for drawing
x
A W xy that is clockwise (positive) on a face resides above the anticlockwise (negative) on a face resides below
x
V axis; a W xy
V axis.
Tensile stress will be positive and plotted right of the origin O. Compressive stress will be negative and will be plotted left to the origin O.
x
An angle
T on real plane transfers as an angle 2T
on Mohrâ€™s circle plane.
We now construct Mohrâ€™s circle in the following stress conditions
I. II.
Biaxial stress when
V x and V y known and W xy = 0
Complex state of stress ( V x ,V y and W xy known)
I. Constant of Mohrâ€™s circle for Biaxial stress (when only V x and If
Vy
known)
V x and V y both are tensile or both compressive sign of V x and V y will be same and this state of
stress is known as â€œ like stressesâ€? if one is tensile and other is compressive sign of
V x and V y will
be opposite and this state of stress is known as â€˜unlike stressâ€™.
x Construction of Mohrâ€™s circle for like stresses (when V x and V y are same type of stress) StepI:
Label the element ABCD and draw all stresses.
StepII: Set up axes for the direct stress (as abscissa) i.e., in xaxis and shear stress (as ordinate) i.e. in Yaxis
Page 68 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
StepIII: Using sign convention and some suitable scale, plot the stresses on two adjacent faces e.g. AB and BC on the graph. Let OL and OM equal to axis O
V.
V x and V y respectively on the
StepIV: Bisect ML at C. With C as centre and CL or CM as radius, draw a circle. It is the Mohrâ€™s circle.
T
StepV: At the centre C draw a line CP at an angle 2 , in the same direction as the normal to the plane makes with the direction of
V x . The point P represents the state of
stress at plane ZB.
Page 69 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
StepVI: Calculation, Draw a perpendicular PQ and PR where PQ = W and PR = V n
OC
Vx Vy
and MC = CL = CP =
Vx Vy
2 PR = V
2
Vx Vy n
Vx Vy
2
PQ = W = CPsin 2T =
cos 2T
2
Vx Vy
sin 2T
2 [Note: In the examination you only draw final figure (which is in StepV) and follow the procedure step by step so that no mistakes occur.]
x Construction of Mohrâ€™s circle for unlike stresses (when V x and V y are opposite in sign) Follow the same steps which we followed for construction for â€˜like stressesâ€™ and finally will get the figure shown below.
Note: For construction of Mohrâ€™s circle for principal stresses when ( V 1 and
V 2 is known) then follow the same steps of Constant of Mohrâ€™s circle for Biaxial stress (when only V x and V y known) just V 1 and V y V 2 change the V x
Page 70 of 429
Chapter2
Principal Stress and Strain
S K Mondal’s
II. Construction of Mohr’s circle for complex state of stress ( V x , V y and W xy known) StepI:
Label the element ABCD and draw all stresses.
StepII: Set up axes for the direct stress (as abscissa) i.e., in xaxis and shear stress (as ordinate) i.e. in Yaxis
StepIII: Using sign convention and some suitable scale, plot the stresses on two adjacent faces
V x and V y respectively on the axis O V . Draw LS perpendicular to oV axis and equal to W xy .i.e. LS=W xy . Here LS is downward as W xy on AB face is (– ive) and draw MT perpendicular to oV axis and equal to W xy i.e. MT=W xy . Here MT is upward as W xy BC face is (+ ive). e.g. AB and BC on the graph. Let OL and OM equal to
Page 71 of 429
Chapter2
Principal Stress and Strain
S K Mondal’s
StepIV: Join ST and it will cut oσ axis at C. With C as centre and CS or CT as radius, draw circle. It is the Mohr’s circle.
StepV: At the centre draw a line CP at an angle 2θ in the same direction as the normal to the plane makes with the direction of σ x .
StepVI: Calculation, Draw a perpendicular PQ and PR where PQ = τ and PR = σ n Centre, OC =
σx +σy 2
2 ⎛ σ x −σ y ⎞ = ⎜ Radius CS = ( CL ) + (LS ) ⎟⎟ +τ xy 2 = CT = CP ⎜ 2 ⎝ ⎠ σ x +σ y σ x −σ y + PR = σ = cos 2θ + τ sin 2θ n xy 2 2 σ x −σ y PQ = τ = sin 2θ −τ xy cos 2θ . 2 2
2
[Note: In the examination you only draw final figure (which is in StepV) and follow the Page 72 of 429
procedure step by step so that no mistakes occur.]
PDF created with pdfFactory Pro trial version www.pdffactory.com
Chapter2
Principal Stress and Strain
Note: The intersections of
oV
S K Mondal’s
axis are two principal stresses, as shown below.
Let us take an example: See the in the Conventional question answer section in this chapter and the question is “Conventional Question IES2000”
2.9 Mohr's circle for some special cases: i) Mohr’s circle for axial loading:
Vx
P ; Vy A
W xy
0
ii) Mohr’s circle for torsional loading:
W xy
Tr ; Vx J
Vy
0
It is a case of pure shear iii) In the case of pure shear
(ǔ1 =  ǔ2 and ǔ3 =Page 0) 73 of 429
Chapter2
Principal Stress and Strain
Vx
V y
W max
rV x
iv) A shaft compressed all round by a hub
ǔ1 = ǔ2 = ǔ3 = Compressive (Pressure) v) Thin spherical shell under internal pressure
V1
V2
pr 2t
pD (tensile) 4t
vi) Thin cylinder under pressure
V1
pD 2t
pr (tensile) and V 2 t
pd 4t
pr (tensile) 2t
vii) Bending moment applied at the free end of a cantilever
Only bending stress, V 1
My and V 2 I
W xy
0 Page 74 of 429
S K Mondal’s
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
2.10 Strain Normal strain Let us consider an element AB of infinitesimal length Ç…x. After deformation of the actual body if displacement of end A is u, that of end B is u+
Â§ ÂŠ
is Â¨ u+
wu Âˇ .G x  u Â¸ wx Âš
wu .G x. This gives an increase in length of element AB wx
wu G x and therefore the strain in xdirection is H x wx
wQ and H z wx
Similarly, strains in y and z directions are H y
wu wx
ww . wz
Therefore, we may write the three normal strain components
Hx
wu ; wx
wQ ; wy
Hy
and
Hz
ww . wz
Change in length of an infinitesimal element.
Shear strain Let us consider an element ABCD in xy plane and let the displaced position of the element be
AcBcCcDc .This gives shear strain in xy plane as J xy v E where v is the angle made by the displaced live BcCc with the vertical and E is the angle made by the displaced line A cDc with the
wu .G y wx Gy
horizontal. This gives v
wQ .G x wu and E = wx Gx wy
wQ wx
We may therefore write the three shear strain components as
J xy
wu wQ ; wy wx
J yz
wQ ww and J zx wz wy
ww wu wx wz
Therefore the state of strain at a point can be completely described by the six strain components and the strain components in their turns can be completely defined by the displacement components u,Q , and w. Therefore, the complete strain matrix can be written as
Page 75 of 429
Chapter2
Principal Stress and Strain ÂŞw Âş 0 0Âť ÂŤ wx ÂŤ Âť w ÂŤ0 0Âť ÂH x Â˝ ÂŤ Âť y w Â° Â° ÂŤ Âť H Â° y Â° ÂŤ w Âť Âu Â˝ 0 Â°H Â° ÂŤ0 wz Âť Â° Â° Â° z Â° Âť ÂŽX Âž ÂŽ Âž ÂŤ w Â°J xy Â° ÂŤ w 0 Âť Â°ÂŻ w Â°Âż Âť Â°J Â° ÂŤ wx wy Âť Â° yz Â° ÂŤ w wÂť Â°ÂŻJ zx Â°Âż ÂŤ0 ÂŤ wy wz Âť ÂŤ Âť w Âť ÂŤw 0 ÂŤÂŹ wz wx ÂťÂź
S K Mondalâ€™s
Shear strain associated with the distortion of an infinitesimal element.
Strain Tensor The three normal strain components are
H x H xx
wu ; wx
Hy
H yy
wQ wy
and
Hz
H zz
ww . wz
The three shear strain components are
Â?xy
J xy 2
1 Â§ wu wQ Âˇ Â¨ Â¸; 2 ÂŠ wy wx Âš
Â?yz
J yz 2
1 Â§ wQ ww Âˇ Â¨ Â¸ 2 ÂŠ wz wy Âš
and
Â?zx
J zx 2
1 Â§ wu ww Âˇ 2 Â¨ÂŠ wz wx Â¸Âš
Therefore the strain tensor is
Â?ij
ÂŞÂ?xx Â?xy ÂŤ ÂŤÂ?yx Â?yy ÂŤ ÂŹÂ?zx Â?zy
Â?xz Âş Âť Â?yz Âť Âť Â?zz Âź
ÂŞ ÂŤÂ?xx ÂŤ ÂŤ J yx ÂŤ ÂŤ 2 ÂŤ J zx ÂŤ ÂŹ 2
J xy 2 Â?yy
J zy 2
J xz Âş
Âť 2 Âť J yz Âť Âť 2 Âť Âť Â?zz Âť Âź
Constitutive Equation The constitutive equations relate stresses and strains and in linear elasticity. We know from the Hookâ€™s law V
E.H
Where E is modulus of elasticity
Page 76 of 429
Chapter2
Principal Stress and Strain
It is known that V x produces a strain of and Poissonâ€™s effect gives P
Vx E
Vx
S K Mondalâ€™s
in xdirection
E
in ydirection and
P
Vx E
in zdirection.
Therefore we my write the generalized Hookâ€™s law as
1 1 ÂŞV y P V z V x Âş and Â?z ÂŞV z P V x V y Âş ÂŹ Âź Âź E EÂŹ It is also known that the shear stress, W GJ , where G is the shear modulus and J is shear strain. Â?x
1 ÂŞV x P V y V z Âş , Âź EÂŹ
Â?y
We may thus write the three strain components as
W xy
J xy
G
J yz
,
W yz G
and J zx
W zx G
In general each strain is dependent on each stress and we may write
ÂH x Â˝ Â° Â° Â°H y Â° Â°H Â° Â° z Â° ÂŽ Âž Â°J xy Â° Â°J Â° Â° yz Â° Â°ÂŻJ zx Â°Âż
?
ÂŞK11 K12 K13 K14 K15 K16 Âş ÂV x Â˝ ÂŤ Âť Â°V Â° ÂŤK 21 K 22 K 23 K 24 K 25 K 26 Âť Â° y Â° ÂŤK 31 K 32 K 33 K 34 K 35 K 36 Âť Â°Â°V z Â°Â° ÂŤ ÂťÂŽ Âž ÂŤK 41 K 42 K 43 K 44 K 45 K 46 Âť Â°W xy Â° ÂŤK K K K K K Âť Â° Â° ÂŤ 51 52 53 54 55 56 Âť Â°W yz Â° ÂŤÂŹK 61 K 62 K 63 K 64 K 65 K 66 ÂťÂź Â°W Â° ÂŻ zx Âż
The number of elastic constant is 36 (For anisotropic materials) For isotropic material
K11 K 22
K 33
K 44
K 55
K 66
K12
K13
K 21
1 E 1 G K 23
K 31
K 32
P E
Rest of all elements in K matrix are zero. For isotropic material only two independent elastic constant is there say E and G.
x 1D Strain Let us take an example: A rod of cross sectional area Ao is loaded by a tensile force P. Itâ€™s stresses
Vx
P , Ao
Vy
0,
and V z
0
1D state of stress or Uniaxial state of stress
V ij
Â§ V xx Â¨ Â¨ 0 Â¨ 0 ÂŠ
0 0Âˇ Â¸ 0 0 Â¸ or W ij 0 0 Â¸Âš
Â§ W xx Â¨ Â¨ 0 Â¨ 0 ÂŠ
0 0Âˇ Â¸ 0 0Â¸ 0 0 Â¸Âš
Â§V x Â¨ Â¨0 Â¨0 ÂŠ
0 0Âˇ Â¸ 0 0Â¸ 0 0 Â¸Âš
Therefore strain components are Page 77 of 429
Chapter2
Â?x
Principal Stress and Strain
Vx E
; Â?y P
Vx E
P Â?x ;
P and Â?z
Vx E
S K Mondalâ€™s
P Â?x
1D state of strain or Uniaxial state of strain
Â§Hx Â¨ Â¨0 Â¨ ÂŠ0
H ij
0 PH x 0
x 2D Strain (V z (i)
Â?x Â?y Â?z
0 Âˇ Â¸ 0 Â¸ PH x ÂšÂ¸
Â§Vx Â¨E Â¨ Â¨ Â¨ 0 Â¨ Â¨ 0 Â¨ ÂŠ
0 P
Vx
0
Âˇ Â¸ Â¸ Â¸ 0 Â¸ Â¸ V P x Â¸Â¸ E Âš 0
E
Â§ py Â¨ Â¨0 Â¨ ÂŠ0
0 qy 0
0Âˇ Â¸ 0Â¸ qy ÂšÂ¸
0)
1 ÂŞÂŹV x PV y ÂşÂź E 1 ÂŞÂŹV y PV x ÂşÂź E P
ÂŞV x V y ÂşÂź EÂŹ
[Where, Â?x ,Â?y ,Â?z are strain component in X, Y, and Z axis respectively]
(ii)
Vx
E ÂŞÂ? P Â?y ÂşÂź 2 ÂŹ x 1 P
Vy
E ÂŞÂ? P Â?x ÂşÂź 2 ÂŹ y 1 P
x 3D Strain (i)
Â?x Â?y Â?z
(ii)
Vx
1ÂŞ V x P V y V z ÂşÂź EÂŹ 1ÂŞ V y P V z V x ÂşÂź ÂŹ E 1ÂŞ V z P V x V y ÂşÂź EÂŹ E ÂŞ1 P Â?x P Â?y Â?z Âş Âź 1 P 1 2 P
ÂŹ Page 78 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
Vy
E ÂŞ1 P Â?y P Â?z Â?x ÂşÂź 1 P 1 2P ÂŹ
Vz
E ÂŞ1 P Â?z P Â?x Â?y Âş Âź 1 P 1 2P ÂŹ
Let us take an example: At a point in a loaded member, a state of plane stress exists and the
270 u 10 6 ;
strains are H x
Hy
90 u 106
and H xy
360 u 10 6 . If the elastic constants
E, P and G are 200 GPa, 0.25 and 80 GPa respectively. Determine the normal stress V x and V y and the shear stress W xy at the point. Answer: We know that
Hx . Hy
H xy This gives V x
and V y
and W xy
1 ^V x PV y ` E 1 ^V y PV x ` E
W xy G
E ^H x PH y ` 1 P 2
200 u 109 ÂŞ 270 u 106 0.25 u 90 u 106 ÂşÂź Pa 1 0.252 ÂŹ 52.8 MPa (i.e. tensile)
E ÂŞH y PH x ÂşÂź 1 P 2 ÂŹ 200 u 109 ÂŞ 90 u 106 0.25 u 270 u 106 ÂşÂź Pa 4.8 MPa (i.e.compressive) 1 0.252 ÂŹ H xy .G 360 u 106 u 80 u 109 Pa 28.8MPa
2.12 An element subjected to strain components Â?x ,Â?y &
J xy 2
Consider an element as shown in the figure given. The strain component In Xdirection is Â?x , the strain component in Ydirection is Â?y and the shear strain component is J xy . Now consider a plane at an angle T with X axis in this plane a normal strain Â?T and a shear strain J T . Then
x
x
Â?T
JT 2
Â?x Â?y 2
Â?x Â?y 2
Â?x Â?y 2 sin 2T
cos 2T
J xy 2
J xy 2
sin 2T
cos 2T Page 79 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
We may find principal strain and principal plane for strains in the same process which we followed for stress analysis. In the principal plane shear strain is zero. Therefore principal strains are
Â?x Â?y
Â?1,2
2
2
Â§ Â?x Â?y Âˇ Â§ J xy Âˇ r Â¨ Â¸ Â¨ Â¸ 2 2 Âš ÂŠ Âš ÂŠ
2
The angle of principal plane
tan 2T p x
J xy (Â?x Â?y )
Maximum shearing strain is equal to the difference between the 2 principal strains i.e
(J xy ) max Â?1 Â?2 Mohr's Circle for circle for Plain Strain We may draw Mohrâ€™s circle for strain following same procedure which we followed for drawing Mohrâ€™s circle in stress. Everything will be same and in the place of
V y write Â?y
and in place of W xy
write
J xy 2
.
Page 80 of 429
V x write Â?x , the place of
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
2.15 Volumetric Strain (Dilation) x
Rectangular block,
'V V0
Â?x Â?y Â?z
Proof: Volumetric strain
V Vo V0
'V V0
L 1 H x u L 1 H y u L 1 H z L3 L3
Before deformation,
Â?x Â?y Â?z
Volume (V)
Volume (Vo) = L3
= L 1 H x u L 1 H y u L 1 H z
(neglecting second and third order term, as very small )
x
In case of prismatic bar,
Volumetric strain,
dv v
H 1 2P
Proof: Before deformation, the volume of the bar, V = A.L After deformation, the length Lc
L 1 H
and the new crosssectional area A c
Therefore now volume V c
After deformation,
A 1 PH
2
A cLc=AL 1 H 1 PH
Page 81 of 429
2
Chapter2
'V V cV V V 'V H 1 2P
V
?
x
Principal Stress and Strain 2 AL 1 H 1 PH AL H 1 2P
AL
S K Mondalâ€™s
Thin Cylindrical vessel
Â? 1=Longitudinal strain =
V1 E
Â?2 =Circumferential strain =
P V2 E
V2 E
P
V1 E
pr >1 2P @ 2 Et pr >2 P @ 2Et
'V pr [5 4Č?] Â?1 2 Â?2 2 Et Vo
x
x
Thin Spherical vessels
Â? Â?1 Â?2
pr [1 P ] 2 Et
'V V0
3 pr [1 P ] 2 Et
3Â?
In case of pure shear
Vx
V y
W
Therefore
Hx
W E
Hy
Hz
0
Therefore H v
dv v
1 P
W E
1 P
Hx Hy Hz
0
2.16 Measurement of Strain Unlike stress, strain can be measured directly. The most common way of measuring strain is by use of the Strain Gauge.
Strain Gauge
Page 82 of 429
Chapter2
Principal Stress and Strain
S K Mondal’s
A strain gage is a simple device, comprising of a thin electric wire attached to an insulating thin backing material such as a bakelite foil. The foil is exposed to the surface of the specimen on which the strain is to be measured. The thin epoxy layer bonds the gauge to the surface and forces the gauge to shorten or elongate as if it were part of the specimen being strained. A change in length of the gauge due to longitudinal strain creates a proportional change in the electric resistance, and since a constant current is maintained in the gauge, a proportional change in voltage. (V = IR). The voltage can be easily measured, and through calibration, transformed into the change in length of the original gauge length, i.e. the longitudinal strain along the gauge length.
Strain Gauge factor (G.F)
The strain gauge factor relates a change in resistance with strain.
Strain Rosette The strain rosette is a device used to measure the state of strain at a point in a plane. It comprises three or more independent strain gauges, each of which is used to read normal strain at the same point but in a different direction. The relative orientation between the three gauges is known as D , E and G The three measurements of normal strain provide sufficient information for the determination of the complete state of strain at the measured point in 2D. We have to find out x , y , and J xy form measured value a , b , and c
Page 83 of 429
Chapter2 General arrangement:
Principal Stress and Strain
S K Mondalâ€™s
The orientation of strain gauges is given in the figure. To relate strain we have to use the following formula.
Â?T
Â?x Â?y 2
Â?x Â?y 2
cos 2T
J xy 2
sin 2T
We get
Â?a
Â?b Â?c
Â?x Â?y 2 Â?x Â?y 2 Â?x Â?y 2
Â?x Â?y 2 Â?x Â?y 2 Â?x Â?y 2
cos 2D
J xy 2
sin 2D
cos 2 D E
J xy 2
cos 2 D E G
sin 2 D E
J xy 2
sin 2 D E G
From this three equations and three unknown we may solve Â?x , Â?y , and J xy
x Two standard arrangement of the of the strain rosette are as follows: (i)
45Â° strain rosette or Rectangular strain rosette. In the general arrangement above, put
D
0o ; E
45o and G
45o
Putting the value we get
x Â?a Â?x Â?x Â?x J xy 2 2 Â?c Â?y
x Â?b x (ii)
45
o
60Â° strain rosette or Delta strain rosette In the general arrangement above, put
D
0o ; E
60o and G
60o
Putting the value we get x Â?a Â?y x
Â?b
x
Â?c
60o
Â?x 3 Â?y
3 J xy 4 4 Â?x 3 Â?y 3 J xy 4 4
or
Solving above three equation we get
Page 84 of 429
1200
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20Years GATE Questions Stresses due to Pure Shear GATE1.
A block of steel is loaded by a tangential force on its top surface while the bottom surface is held rigidly. The deformation of the block is due to [GATE1992] (a) Shear only (b) Bending only (c) Shear and bending (d) Torsion GATE1. Ans. (a) It is the definition of shear stress. The force is applied tangentially it is not a point load so you cannot compare it with a cantilever with a point load at its free end. A shaft subjected to torsion experiences a pure shear stress W on the surface. The maximum principal stress on the surface which is at 45Â° to the axis will have a value [GATE2003] (b) 2 W cos 45Â° (c) W cos2 45Â° (d) 2 W sin 45Â° cos 45Â° (a) W cos 45Â° Vx Vy Vx Vy GATE2. Ans. (d) V n cos 2T W xy sin 2T 2 2 Here V x V 2 0, W xy W , T 45o
GATE2.
GATE3.
The number of components in a stress tensor defining stress at a point in three dimensions is: [GATE2002] (a) 3 (b) 4 (c) 6 (d) 9 GATE3. Ans. (d) It is well known that,
W xy
W yx, W xz
W zx and W yz
W zy
so that the state of stress at a point is given by six components V x ,V y ,V z and W xy , W yz ,W zx
Principal Stress and Principal Plane GATE4.
A body is subjected to a pure tensile stress of 100 units. What is the maximum shear produced in the body at some oblique plane due to the above? [IES2006] (a) 100 units (b) 75 units (c) 50 units (d) 0 unit V 1 V 2 100 0 50 units. GATE4. Ans. (c) W max 2 2 GATE5.
In a strained material one of the principal stresses is twice the other. The maximum shear stress in the same case is W max .Then, what is the value of the maximum principle stress? (b) 2 W max (a) W max
GATE5. Ans. (c) GATE6.
W max
V1 V 2 2
,
V1
2V 2 or W max
(c) 4 W max
V2 2
or
V2
[IES 2007] (d) 8 W max
2W max or V 1 2V 2 = 4W max
A material element subjected to a plane state of stress such that the maximum shear stress is equal to the maximum tensile stress, would correspond to [IAS1998]
Page 85 of 429
Chapter2
Principal Stress and Strain
GATE6. Ans. (d) W max GATE7.
V 1 ( V 1 )
2
2
V1
A solid circular shaft is subjected to a maximum shearing stress of 140 MPs. The magnitude of the maximum normal stress developed in the shaft is: [IAS1995] (a) 140 MPa (b) 80 MPa (c) 70 MPa (d) 60 MPa
GATE7. Ans. (a) W max GATE8.
V1 V 2
V1 V 2 2
Maximum normal stress will developed if V 1
V 2
V
The state of stress at a point in a loaded member is shown in the figure. The [IAS 1994] magnitude of maximum shear stress is [1MPa = 10 kg/cm2] (a) 10 MPa (b) 30 MPa (c) 50 MPa (d) 100MPa
2
GATE8. Ans. (c)
GATE9.
S K Mondalâ€™s
W max
2
Â§Vx V y Â· Â¨Â¨ Â¸Â¸ W xy 2 = Â© 2 Â¹
Â§ 40 40 Â· 2 Â¨ Â¸ 30 = 50 MPa 2 Â© Â¹
A solid circular shaft of diameter 100 mm is subjected to an axial stress of 50 MPa. It is further subjected to a torque of 10 kNm. The maximum principal stress experienced on the shaft is closest to [GATE2008] (a) 41 MPa (b) 82 MPa (c) 164 MPa (d) 204 MPa
GATE9. Ans. (b) Shear Stress ( W )=
16 u 10000 Pa S u (0.1)3
16T Sd 3
Maximum principal Stress =
50.93 MPa
2
Vb
Â§V Â· Â¨ b Â¸ W 2 =82 MPa 2 Â© 2 Â¹
GATE10. In a biaxial stress problem, the stresses in x and y directions are (Ç”x = 200 MPa [GATE2000] and Ç”y =100 MPa. The maximum principal stress in MPa, is: (a) 50 (b) 100 (c) 150 (d) 200 GATE10. Ans. (d) V 1
Vx Vy 2
2
Â§Vx Vy Â· 2 Â¨ Â¸ W xy 2 Â© Â¹
if W xy
Page 86 of 429
0
Chapter2
Principal Stress and Strain Vx Vy 2
Â§Vx Vy Â· Â¨ Â¸ 2 Â¹ Â©
S K Mondalâ€™s
2
Vx
GATE11. The maximum principle stress for the stress state shown in the figure is (a) Ç” (b) 2 Ç” (c) 3 Ç” (d) 1.5 Ç”
[GATE2001] GATE11. Ans. (b) V x ?V 1 max
V, Vy
Vx Vy 2
V , W xy
V 2
Â§Vx Vy Â· 2 Â¨ Â¸ W xy 2 Â© Â¹
V V 2
0
2
V2
2V
GATE12. The normal stresses at a point are Ç”x = 10 MPa and, Ç”y = 2 MPa; the shear stress at this point is 4MPa. The maximum principal stress at this point is: [GATE1998] (a) 16 MPa (b) 14 MPa (c) 11 MPa (d) 10 MPa GATE12. Ans. (c) V 1
Vx Vy 2
2
Â§Vx Vy Â· 2 Â¨ Â¸ W xy 2 Â© Â¹
2
10 2 Â§ 10 2 Â· 2 Â¨ Â¸ 4 2 2 Â© Â¹
GATE13. In a Mohr's circle, the radius of the circle is taken as: 2
(a)
2 Â§ V x V y Â· Â¨ Â¸ W xy
Â© 2 Â¹
(c)
2 Â§ V x V y Â· Â¨ Â¸ W xy
Â© 2 Â¹
(b)
V
x
V y
11.66 MPa
[IES2006; GATE1993]
2
2
W xy
2
2
(d)
V
2
x
V y W xy
2
Where, Ç”x and Ç”y are normal stresses along x and y directions respectively and Ç•xy is the shear stress. GATE13. Ans. (a)
GATE14. A two dimensional fluid element rotates like a rigid body. At a point within the element, the pressure is 1 unit. Radius of the Mohr's circle, characterizing the state of stress at that point, is: [GATE2008] (a) 0.5 unit (b) 0 unit (c) 1 unit (d) 2 units GATE14. Ans. (b)
Page 87 of 429
Chapter2
Principal Stress and Strain
GATE15. The Mohr's circle of plane stress for a point in a body is shown. The design is to be done on the basis of the maximum shear stress theory for yielding. Then, yielding will just begin if the designer chooses a ductile material whose yield strength is: (a) 45 MPa (b) 50 MPa (c) 90 MPa (d) 100 MPa GATE15. Ans. (c) Given V 1
10 MPa,
V2
Vy
Â&#x; Vy
[GATE2005]
100 MPa
Maximum shear stress theory give W max or V 1 V 2
S K Mondalâ€™s
V1 V 2
10 ( 100)
Vy
2 2 90MPa
GATE16. The figure shows the state of stress at a certain point in a stressed body. The magnitudes of normal stresses in the x and y direction are 100MPa and 20 MPa respectively. The radius of Mohr's stress circle representing this state of stress is: (a) 120 (b) 80 (c) 60 (d) 40 [GATE2004] GATE16. Ans. (c) V x 100MPa, V y
20MPa
Radius of Mohr 's circle
Vx Vy
100 20
2
2
60
Data for Q17â€“Q18 are given below. Solve the problems and choose correct answers. [GATE2003] The state of stress at a point "P" in a two dimensional loading is such that the Mohr's circle is a point located at 175 MPa on the positive normal stress axis. GATE17. Determine the maximum and minimum principal stresses respectively from the Mohr's circle (a) + 175 MPa, â€“175MPa (b) +175 MPa, +175 MPa (c) 0, â€“175 MPa (d) 0, 0 GATE17. Ans. (b)
V1 V 2
Vx
Vy
175 MPa
GATE18. Determine the directions of maximum and minimum principal stresses at the point â€œPâ€? from the Mohr's circle Page 88 of 429 [GATE2003]
Chapter2
Principal Stress and Strain
(a) 0, 90Â° (b) 90Â°, 0 (c) 45Â°, 135Â° GATE18. Ans. (d) From the Mohrâ€™s circle it will give all directions.
S K Mondalâ€™s (d) All directions
Principal strains GATE19. If the two principal strains at a point are 1000 Ă— 106 and 600 Ă— 106, then the maximum shear strain is: [GATE1996] (b) 500 Ă— 106 (c) 1600 Ă— 106 (d) 200 Ă— 106 (a) 800 Ă— 106 GATE19. Ans. (c) Shear strain emax emin ^1000 600 ` u 10 6 1600 u 10 6
Previous 20Years IES Questions Stresses due to Pure Shear IES1.
If a prismatic bar be subjected to an axial tensile stress Ç”, then shear stress induced on a plane inclined at Ç‰ with the axis will be: [IES1992]
a
V 2
sin 2T
b
V 2
cos 2T
c
V 2
d
cos 2 T
V 2
sin 2 T
IES1. Ans. (a) IES2.
In the case of biaxial state of normal stresses, the normal stress on 45Â° plane is equal to [IES1992] (a) The sum of the normal stresses (b) Difference of the normal stresses (c) Half the sum of the normal stresses (d) Half the difference of the normal stresses Vx Vy Vx Vy IES2. Ans. (c) V n cos 2T W xy sin 2T 2 2 Vx Vy At T 45o andW xy 0; V n 2 IES3.
In a twodimensional problem, the state of pure shear at a point is characterized by [IES2001] (b) H x H y and J xy z 0 (a) H x H y and J xy 0 (c)
Hx
2H y and J xy z 0
(d)
Hx
0.5H y and J xy
0
IES3. Ans. (b) IES4.
Which one of the following Mohrâ€™s circles represents the state of pure shear? [IES2000]
IES4. Ans. (c) Page 89 of 429
Chapter2 IES5.
Principal Stress and Strain
S K Mondalâ€™s
For the state of stress of pure shear W the strain energy stored per unit volume in the elastic, homogeneous isotropic material having elastic constants E and [IES1998] Q will be:
W2
1 Q
E IES5. Ans. (a) V 1 W , V2 (a)
U
W2
2W 2 (c) 1 Q
E
1 Q
2E W , V 3 0 (b)
1 ÂŞ 2 2 W W 2PW W Âş V ÂŹ Âź 2E
(d)
W2 2E
2 Q
1 P 2 W V E
IES6.
Assertion (A): If the state at a point is pure shear, then the principal planes through that point making an angle of 45Â° with plane of shearing stress carries principal stresses whose magnitude is equal to that of shearing stress. Reason (R): Complementary shear stresses are equal in magnitude, but opposite in direction. [IES1996] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES6. Ans. (b) IES7.
Assertion (A): Circular shafts made of brittle material fail along a helicoidally surface inclined at 45Â° to the axis (artery point) when subjected to twisting moment. [IES1995] Reason (R): The state of pure shear caused by torsion of the shaft is equivalent to one of tension at 45Â° to the shaft axis and equal compression in the perpendicular direction. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES7. Ans. (a) Both A and R are true and R is correct explanation for A. IES8.
A state of pure shear in a biaxial state of stress is given by
Â§V1 0 Âˇ Â¸ ÂŠ 0 V2 Âš IES8. Ans. (b) V 1 W , V2
Â§V1 0 Âˇ Â¸ ÂŠ 0 V 1 Âš W , V 3 0 (b) Â¨
(a) Â¨
(c)
Â§ V x W xy Âˇ Â¨ Â¸ ÂŠW yx V y Âš
[IES1994]
(d) None of the above
IES9.
The state of plane stress in a plate of 100 mm thickness is given as [IES2000] Ç”xx = 100 N/mm2, Ç”yy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is: (c) 100 N/mm2 (d) 200 N/mm2 (a) Zero (b) 90 N/mm2 IES9. Ans. (a) IES10.
The state of plane stress at a point is described by
V x V y V and W xy
normal stress on the plane inclined at 45Â° to the xplane will be:
a V IES10. Ans. (a) V n IES11.
b
Vx Vy 2
Vx Vy 2
2V
c
3V
0 . The
[IES1998]
d 2V
cos 2T W xy sin 2T
Consider the following statements: [IES1996, 1998] State of stress in two dimensions at a point in a loaded component can be completely specified by indicating the normal and shear stresses on 1. A plane containing the point 2. Any two planes passing through the point Page 90 of 429 3. Two mutually perpendicular planes passing through the point
Chapter2
Principal Stress and Strain
Of these statements (a) 1, and 3 are correct (c) 1 alone is correct IES11. Ans. (d)
S K Mondalâ€™s
(b) 2 alone is correct (d) 3 alone is correct
Principal Stress and Principal Plane IES12.
A body is subjected to a pure tensile stress of 100 units. What is the maximum shear produced in the body at some oblique plane due to the above? [IES2006] (a) 100 units (b) 75 units (c) 50 units (d) 0 unit V 1 V 2 100 0 50 units. IES12. Ans. (c) W max 2 2 IES13.
In a strained material one of the principal stresses is twice the other. The maximum shear stress in the same case is U max . Then, what is the value of the maximum principle stress? [IES 2007] W W W (a) max (b) 2 max (c) 4 max (d) 8 W max
IES13. Ans. (c) IES14.
V1 V 2
W max
2
V x uV y
W xy
if V 2
(b)
2 Vx Vy
IES14. Ans. (c) V 1,2
2
V2 2
or
V2
2W max or V 1 2V 2 = 4W max
W xy
V x uV y
(c)
W xy
V x uV y
(d)
W xy
V x2 V y2
2
Â§Vx Vy Â· 2 Â¨ Â¸ W xy 2 Â© Â¹ 2
Vx Vy
Â&#x;
0
Â§Vx Vy Â· 2 Â¨ Â¸ W xy 2 Â© Â¹
2
Â§Vx Vy Â· or Â¨ Â¸ 2 Â¹ Â©
2
2
Â§Vx Vy Â· 2 Â¨ Â¸ W xy or W xy 2 Â© Â¹
Vx uVy
The principal stresses Ç”1, Ç”2 and Ç”3 at a point respectively are 80 MPa, 30 MPa and â€“40 MPa. The maximum shear stress is: [IES2001] (a) 25 MPa (b) 35 MPa (c) 55 MPa (d) 60 MPa
IES15. Ans. (d) W max
IES16.
2V 2 or W max
V1
In a strained material, normal stresses on two mutually perpendicular planes are Ç”x and Ç”y (both alike) accompanied by a shear stress Ç•xy One of the principal stresses will be zero, only if [IES2006] (a)
IES15.
,
V1 V 2 2
80 ( 40) 2
60 MPa
Plane stress at a point in a body is defined by principal stresses 3Ç” and Ç”. The ratio of the normal stress to the maximum shear stresses on the plane of maximum shear stress is: [IES2000] (a) 1 (b) 2 (c) 3 (d) 4
IES16. Ans. (b) tan 2T
W max
V1 V 2 2
2W xy
Vx Vy
Â&#x;T
3V V 2
0
V
Major principal stress on the plane of maximum shear = V 1 Page 91 of 429
3V V 2
2V
Chapter2 IES17.
Principal Stress and Strain
S K Mondalâ€™s
Principal stresses at a point in plane stressed element are
Vx
Vy
500 kg/cm 2 .
Normal stress on the plane inclined at 45o to xaxis will be: [IES1993] 2 2 (a) 0 (b) 500 kg/cm (c) 707 kg/cm (d) 1000 kg/cm2 IES17. Ans. (b) When stresses are alike, then normal stress Ç”n on plane inclined at angle 45Â° is 2
2
Â§ 1 Âˇ Â§ 1 Âˇ Vy Â¨ Â¸ Vx Â¨ Â¸ ÂŠ 2Âš ÂŠ 2Âš
2
V n V y cos T V x sin T IES18.
2
ÂŞ1 1Âş 500 ÂŤ Âť ÂŹ2 2Âź
500 kg/cm 2
If the principal stresses corresponding to a twodimensional state of stress are
V1
V2
and
is greater than
V2
and both are tensile, then which one of the
following would be the correct criterion for failure by yielding, according to the maximum shear stress criterion? [IES1993]
(a)
V 1 V 2
r
2
V yp 2
(b)
V1
r
2
V yp 2
(c )
V2 2
r
V yp 2
(d ) V 1
r2V yp
IES18. Ans. (a) IES19.
For the state of plane stress. Shown the maximum and minimum principal stresses are: (a) 60 MPa and 30 MPa (b) 50 MPa and 10 MPa (c) 40 MPa and 20 MPa (d) 70 MPa and 30 MPa [IES1992]
IES19. Ans. (d) V 1,2
Vx Vy 2
2
Â§Vx Vy Âˇ 2 Â¨ Â¸ W xy 2 ÂŠ Âš 2
V 1,2 V max
50 ( 10) Â§ 50 10 Âˇ 2 Â¨ Â¸ 40 2 ÂŠ 2 Âš 70 and V min
30
IES20.
Normal stresses of equal magnitude p, but of opposite signs, act at a point of a strained material in perpendicular direction. What is the magnitude of the resultant normal stress on a plane inclined at 45Â° to the applied stresses? [IES2005] (a) 2 p (b) p/2 (c) p/4 (d) Zero Vx Vy Vx Vy IES20. Ans. (d) V x cos 2T 2 2 P P P P cos 2 u 45 0 Vn 2 2 IES21.
A plane stressed element is subjected to the state of stress given by
V x W xy 100 kgf/cm 2 and Ç”y = 0. Maximum shear stress in the element is equal to
a 50
[IES1997]
3 kgf/cm 2
b 100 kgf/cm 2 Page 92 of 429
c 50
5 kgf/cm2
d 150 kgf/cm 2
Chapter2
Principal Stress and Strain
IES21. Ans. (c)
Vx 0
V 1,2
2
Â§V 0Â· r Â¨ x W xy2 Â¸ Â© 2 Â¹
Maximum shear stress =
IES22.
S K Mondalâ€™s
2
V 1 V 2 2
50 # 50 5
50 5
Match List I with List II and select the correct answer, using the codes given below the lists: [IES1995] List I(State of stress) List II(Kind of loading)
Codes: (a) (c) IES22. Ans. (c)
A 1 2
B 2 4
C 3 3
D 4 1
A 2 3
(b) (d)
B 3 4
C 4 1
D 1 2
Mohr's circle IES23.
Consider the Mohr's circle shown above: What is the state of stress represented by this circle? (a)V x V y z 0, W xy 0
(b)V x V y (c)V x
0, W xy z 0
0, V y W xy z 0
(d)V x z 0, V y
W xy
0 [IES2008]
IES23. Ans. (b) It is a case of pure shear. Just put V 1 IES24.
V 2
For a general two dimensional stress system, what are the coordinates of the centre of Mohrâ€™s circle? (a)
Vx V y 2
,0
(b) 0,
Vx Vy 2
(c)
Vx Vy 2
,0
(d) 0,
Vx V y 2
IES24. Ans. (c) IES25.
In a Mohr's circle, the radius of the circle is taken as: 2
(a)
2 Â§ V x V y Â· Â¨ Â¸ W xy
Â© 2 Â¹
(c)
2 Â§ V x V y Â· Â¨ Â¸ W xy
Â© 2 Â¹
(b)
V
x
[IES2006; GATE1993]
V y
2
2
W xy
2
2
Page 93 of 429
(d)
V
2
x
V y W xy
2
[IE
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
Where, Ç”x and Ç”y are normal stresses along x and y directions respectively and Ç•xy is the shear stress. IES25. Ans. (a)
IES26.
Maximum shear stress in a Mohr's Circle [IES 2008] (a) Is equal to radius of Mohr's circle (b) Is greater than radius of Mohr's circle (c) Is less than radius of Mohr's circle (d) Could be any of the above IES26. Ans. (a)
Â§ Â¨ Â§ Vx Vy Â¨ Â¨Â¨ 2 Â¨ Â© Â© ? Radius of the Mohr Circle Vx Vy Â§ Â¨Â¨ V xc 2 Â©
2
Â· 2 Â¸Â¸ Wxcyc Â¹
2 Â· Â· 2 Â¸ Â¸Â¸ W xy Â¸ Â¸ Â¹ Â¹
2
2
Â§ Vx Vy Â· 2 Â¨Â¨ Â¸Â¸ W xy 2 Â© Â¹ Vx Vy
? Vt
V2
2 Vx Vy
Â&#x; Wmax
IES27.
2
Â§ Vx Vy Â¨Â¨ 2 Â©
Â§ Vx Vy Â¨Â¨ 2 Â©
V1 V2 2
r
2
Â· 2 Â¸Â¸ W xy Â¹
2
Â· 2 Â¸Â¸ W xy Â¹ Â&#x; Wmax
Â§ Vx Vy Â¨Â¨ 2 Â©
2
Â· 2 Â¸Â¸ W xy Â¹
At a point in twodimensional stress system Ç”x = 100 N/mm2, Ç”y = Ç•xy = 40 N/mm2. What is the radius of the Mohr circle for stress drawn with a scale of: 1 cm = 10 [IES2005] N/mm2? Page 94 of 429 (a) 3 cm (b) 4 cm (c) 5 cm (d) 6 cm
Chapter2
Principal Stress and Strain
IES27. Ans. (c) Radius of the Mohr circle ÂŞ V V 2 Âş ÂŞ Â§ 100 40 Âˇ2 Âş y Âˇ 2 ÂŤ Â¨Â§ x ÂŤ Â¨ Âť / 10 40 W xy 2 Âť / 10 Â¸ Â¸ ÂŤ ÂŠ Âť 2 Âš 2 ÂŤ ÂŠ Âť Âš ÂŹ Âź ÂŹ Âź IES28.
S K Mondalâ€™s
50 / 10
5 cm
Consider a two dimensional state of stress given for an element as shown in the diagram given below: [IES2004]
What are the coordinates of the centre of Mohr's circle? (a) (0, 0) (b) (100, 200) (c) (200, 100) Â§ V x V y Âˇ Â§ 200 100 Âˇ ,0 Â¸ Â¨ ,0 Â¸ 50,0
IES28. Ans. (d) Centre of Mohrâ€™s circle is Â¨ 2 Âš ÂŠ 2 Âš ÂŠ
(d) (50, 0)
IES29.
Twodimensional state of stress at a point in a plane stressed element is represented by a Mohr circle of zero radius. Then both principal stresses (a) Are equal to zero [IES2003] (b) Are equal to zero and shear stress is also equal to zero (c) Are of equal magnitude but of opposite sign (d) Are of equal magnitude and of same sign IES29. Ans. (d) IES30.
Assertion (A): Mohr's circle of stress can be related to Mohr's circle of strain by some constant of proportionality. [IES2002] Reason (R): The relationship is a function of yield stress of the material. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES30. Ans. (c) IES31.
When two mutually perpendicular principal stresses are unequal but like, the maximum shear stress is represented by [IES1994] (a) The diameter of the Mohr's circle (b) Half the diameter of the Mohr's circle (c) Onethird the diameter of the Mohr's circle (d) Onefourth the diameter of the Mohr's circle IES31. Ans. (b) IES32.
State of stress in a plane element is shown in figure I. Which one of the following figuresII is the correct sketch of Mohr's circle of the state of stress? [IES1993, 1996]
FigureI
Page 95 of 429
FigureII
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
IES32. Ans. (c)
Strain IES33.
A point in a two dimensional state of strain is subjected to pure shearing strain of magnitude J xy radians. Which one of the following is the maximum principal strain? (a) J xy
(b) J xy / 2
[IES2008] (d) 2 J xy
(c) J xy /2
IES33. Ans. (c) IES34.
Assertion (A): A plane state of stress does not necessarily result into a plane state of strain as well. [IES1996] Reason (R): Normal stresses acting along X and Y directions will also result into normal strain along the Zdirection. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES34. Ans. (a)
Principal strains Principal strains at a point are 100 u 106 and 200 u 106 . What is the maximum shear strain at the point? [IES2006] â€“6 â€“6 â€“6 (b) 200 Ă— 10 (c) 150 Ă— 10 (d) 100 Ă— 10â€“6 (a) 300 Ă— 10 6 6 IES35. Ans. (a) J max H 1 H 2 100 200 u 10 300 u 10 IES35.
don' t confuse withMaximumShear stress W max
in strain
J xy
H1 H 2
2
2
and W max
V1 V 2 2
V1 V 2 2
that is the difference.
IES36.
The principal strains at a point in a body, under biaxial state of stress, are 1000Ă—10â€“6 and â€“600 Ă— 10â€“6. What is the maximum shear strain at that point? [IES2009] (b) 800 Ă— 10â€“6 (c) 1000 Ă— 10â€“6 (d) 1600 Ă— 10â€“6 (a) 200 Ă— 10â€“6 IES36. Ans. (d)
Â?x Â?y
Ixy
2
2
Â&#x;
Ixy
Â?x Â?y
1000 u 106 600 u 106
1600 u 106
IES37.
The number of strain readings (using strain gauges) needed on a plane surface to determine the principal strains and their directions is: [IES1994] (a) 1 (b) 2 (c) 3 (d) 4 IES37. Ans. (c) Three strain gauges are needed on a plane surface to determine the principal strains and their directions.
Principal strain induced by principal stress IES38.
The principal stresses at a point in two dimensional stress system are V 1 and V 2 and corresponding principal strains are H1 and H 2 . If E and Q denote Young's modulus and Poisson's ratio, respectively, then which one of the following is correct? [IES2008] E (a) V 1 EH1 (b)V 1 >H1 QH 2 @ 1Q 2 E (c)V 1 (d)96Vof E >H 1 QH 2 @ >H1 QH 2 @ Page 1 429 1Q 2
Chapter2 IES38. Ans. (b) F1
T1 T N 2 E E
Principal Stress and Strain S K Mondalâ€™s T T and F2 2 N 1 From these two equation eliminate T2 . E E
IES39.
Assertion (A): Mohr's construction is possible for stresses, strains and area moment of inertia. [IES2009] Reason (R): Mohr's circle represents the transformation of secondorder tensor. (a) Both A and R are individually true and R is the correct explanation of A. (b) Both A and R are individually true but R is NOT the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true. IES39. Ans. (a)
Previous 20Years IAS Questions Stresses due to Pure Shear IAS1.
On a plane, resultant stress is inclined at an angle of 45o to the plane. If the [IAS2003] normal stress is 100 N /mm2, the shear stress on the plane is: (b) 100 N/mm2 (c) 86.6 N/mm2 (d) 120.8 N/mm2 (a) 71.5 N/mm2 IAS1. Ans. (b) We know V n V cos2 T and W V sin T cos T
100 V cos2 45 or V
W IAS2.
200
200 sin 45 cos 45 100
Biaxial stress system is correctly shown in
[IAS1999]
IAS2. Ans. (c)
IAS3.
The complementary shear stresses of intensity W are induced at a point in the material, as shown in the figure. Which one of the following is the correct set of orientations of principal planes with respect to AB? (a) 30Â° and 120Â° (b) 45Â° and 135Â° (c) 60Â° and 150Â° (d) 75Â°Page and97165Â° of 429 [IAS1998]
Chapter2
Principal Stress and Strain
S K Mondal’s
IAS3. Ans. (b) It is a case of pure shear so principal planes will be along the diagonal. IAS4.
A uniform bar lying in the xdirection is subjected to pure bending. Which one of the following tensors represents the strain variations when bending moment is about the zaxis (p, q and r constants)? [IAS2001]
§ py 0 0 · ¨ ¸ (a) 0 qy 0 ¸ ¨ ¨ 0 0 ry ¸ © ¹ 0 · § py 0 ¨ ¸ (c) 0 py 0 ¸ ¨ ¨ 0 0 py ¸¹ ©
§ py 0 0 · ¨ ¸ (b) 0 qy 0 ¸ ¨ ¨ 0 0 0 ¸¹ © § py 0 0 · ¨ ¸ (d) 0 qy 0 ¨ ¸ ¨ 0 0 qy ¸ © ¹
IAS4. Ans. (d) Stress in x direction = ǔx Therefore IAS5.
Vx , E
Hx
Hy
P
Vx , E
Hz
P
Vx E
Assuming E = 160 GPa and G = 100 GPa for a material, a strain tensor is given as: [IAS2001]
§ 0.002 0.004 0.006 · ¨ ¸ 0 ¸ ¨ 0.004 0.003 ¨ 0.006 0 0 ¸¹ © The shear stress,
W xy is:
(a) 400 MPa
(b) 500 MPa
(c) 800 MPa
(d) 1000 MPa
IAS5. Ans. (c)
W xy
G J xy
ªH xx H xy H xz º J xy « » «H yx H yy H yz » and H xy 2 «H H H » ¬ zx zy zz ¼ 3 100 u 10 u 0.004 u 2 MPa 800MPa
Principal Stress and Principal Plane IAS6.
A material element subjected to a plane state of stress such that the maximum shear stress is equal to the maximum tensile stress, would correspond to [IAS1998]
IAS6. Ans. (d) W max IAS7.
V1 V 2
V 1 ( V 1 )
2
2
V1
A solid circular shaft is subjected to a maximum shearing stress of 140 MPs. The magnitude of the maximum normal stress developed in the shaft is: Page 98 of 429 [IAS1995]
Chapter2
Principal Stress and Strain (a) 140 MPa
IAS7. Ans. (a) W max IAS8.
(b) 80 MPa
V1 V 2 2
Maximum normal stress will developed if V 1
IAS8. Ans. (c)
W max
Â§V x V y Âˇ Â¨Â¨ Â¸Â¸ W xy 2 = 2 ÂŠ Âš
V 2
V
2
Â§ 40 40 Âˇ 2 Â¨ Â¸ 30 = 50 MPa 2 ÂŠ Âš
A horizontal beam under bending has a maximum bending stress of 100 MPa and a maximum shear stress of 20 MPa. What is the maximum principal stress in the beam? [IAS2004]
(a) 20 IAS9. Ans. (c) Ç”b=100MPa Ç”1,2=
(b) 50
(c) 50 +
2900
(d) 100
W =20 mPa 2
Vb
V 1,2 IAS10.
(d) 60 MPa
The state of stress at a point in a loaded member is shown in the figure. The [IAS 1994] magnitude of maximum shear stress is [1MPa = 10 kg/cm2] (a) 10 MPa (b) 30 MPa (c) 50 MPa (d) 100MPa
2
IAS9.
S K Mondalâ€™s
(c) 70 MPa
Â§V Âˇ Â¨ b Â¸ W 2 2 ÂŠ 2 Âš 2
Vb
2
Â§V Âˇ Â¨ b Â¸ W 2 2 ÂŠ 2 Âš
100 Â§ 100 Âˇ 2 Â¨ Â¸ 20 2 ÂŠ 2 Âš
50
2900 MPa
When the two principal stresses are equal and like: the resultant stress on any plane is: [IAS2002] (a) Equal to the principal stress (b) Zero (c) One half the principal stress (d) One third of the principal stress
IAS10. Ans. (a)
Vn
Vx V y 2
V x V y
[We may consider this as
2
W xy
cos 2T 0 ] Vx
V y V ( say )
IAS11.
So
V n V for any plane
Assertion (A): When an isotropic, linearly elastic material is loaded biaxially, the directions of principal stressed are different from those of principal strains. [IAS2001] Reason (R): For an isotropic, linearly elastic material the Hooke's law gives only two independent material properties. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS11. Ans. (d) They are same. Page 99 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
IAS12.
Principal stress at a point in a stressed solid are 400 MPa and 300 MPa respectively. The normal stresses on planes inclined at 45Â° to the principal planes will be: [IAS2000] (a) 200 MPa and 500 MPa (b) 350 MPa on both planes (c) 100MPaand6ooMPa (d) 150 MPa and 550 MPa IAS12. Ans. (b)
Vn
Â§ Vx V y Â¨ ÂŠ 2
Âˇ Â§ V x V y Â¸Â¨ Âš ÂŠ 2
Âˇ Â¸ cos 2T Âš
400 300 400 300 cos 2 u 45o 2 2
350MPa
The principal stresses at a point in an elastic material are 60N/mm2 tensile, 20 N/mm2 tensile and 50 N/mm2 compressive. If the material properties are: Îź = 0.35 and E = 105 Nmm2, then the volumetric strain of the material is: [IAS1997] (b) 3 Ă— 104 (c) 10.5 Ă— 10â€“5 (d) 21 Ă— 10â€“5 (a) 9 Ă— 10â€“5 IAS13. Ans. (a) Vy Vy Âˇ Â§Vy Vz Âˇ Â§V Vx V Âˇ Vz Â§V Â?x PÂ¨ P Â¨ z x Â¸ and Â?z PÂ¨ x Â¸ , Â?y Â¸ E EÂš E E Âš E E Âš ÂŠE ÂŠE ÂŠE
IAS13.
Â?v Â?x Â?y Â?z
Vx Vy Vz E
Â§Vx Vy Vz Âˇ Â¸ E ÂŠ Âš
1 2P Â¨
2P V x V y V z
E
Â§ 60 20 50 Âˇ Â¨ Â¸ 1 2 u 0.35
105 ÂŠ Âš
9 u 10 5
Mohr's circle IAS14.
Match ListI (Mohr's Circles of stress) with ListII (Types of Loading) and select the correct answer using the codes given below the lists: [IAS2004] ListI ListII (Mohr's Circles of Stress) (Types of Loading)
Codes: (a) (c) IAS14. Ans. (d)
A 5 4
B 4 3
C 3 2
D 2 5
1.
A shaft compressed all round by a hub
2.
Bending moment applied at the free end of a cantilever
3.
Shaft under torsion
4.
Thin cylinder under pressure
5.
Thin spherical shell under internal pressure
(b) (d)
Page 100 of 429
A 2 2
B 4 3
C 1 1
D 3 5
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
IAS15.
The resultant stress on a certain plane makes an angle of 20Â° with the normal to the plane. On the plane perpendicular to the above plane, the resultant stress makes an angle of Ç‰ with the normal. The value of Ç‰ can be: [IAS2001] (a) 0Â° or 20Â° (b) Any value other than 0Â° or 90Â° (c) Any value between 0Â° and 20Â° (d) 20Â° only IAS15. Ans. (b) IAS16.
The correct Mohr's stresscircle drawn for a point in a solid shaft compressed by a shrunk fit hub is as (OOrigin and CCentre of circle; OA = Ç”1 and OB = Ç”2) [IAS2001]
IAS16. Ans. (d) IAS17.
A Mohr's stress circle is drawn for a body subjected to tensile stress in two mutually perpendicular directions such that
f x > f y . Which one of the
following statements in this regard is NOT correct?
f x is equal to
(a) Normal stress on a plane at 45Â° to (b) Shear stress on a plane at 450 to
f x is equal to
(c) Maximum normal stress is equal to (d) Maximum shear stress is equal to IAS17. Ans. (d) Maximum shear stress is
IAS18.
fx f y 2 fx f y 2
fx . fx f y 2
2
V x =2
N/mm2,
Mohrâ€™s circle is:
V x =0
and W xy
0 , the correct [IAS1999]
Â§Vx Vy Âˇ ,0 Â¸ IAS18. Ans. (d) Centre Â¨ ÂŠ 2 Âš 2
IAS19.
[IAS2000]
fx f y
For the given stress condition
radius
f x and f y
Â§Vx Vy Âˇ 2 Â¨ Â¸ Wx 2 ÂŠ Âš
Â§20 Âˇ Â¨ 2 ,0 Â¸ ÂŠ Âš
1, 0
2
Â§20Âˇ Â¨ 2 Â¸ 0 ÂŠ Âš
1
For which one of the following twodimensional states of stress will the Mohr's stress circle degenerate into a point? [IAS1996]
Page 101 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
IAS19. Ans. (c) Mohrâ€™s circle will be a point. 2
Radius of the Mohrâ€™s circle =
Â§Vx Vy Âˇ 2 Â¨ Â¸ W xy ÂŠ 2 Âš
? W xy
0 andV x
Vy
V
Principal strains IAS20.
In an axisymmetric plane strain problem, let u be the radial displacement at r. Then the strain components H r , HT , beT are given by [IAS1995] (a)
Hr
(c)
Hr
u , HT r u , HT r
wu , b rT wr wu , b rT wr
w 2u wrwT
(b)
Hr
0
(d)
Hr
wu , HT wr wu , HT wr
u , b rT o r wu w 2u , b rT wT wrwT
IAS20. Ans. (b) IAS21.
Assertion (A): Uniaxial stress normally gives rise to triaxial strain. Reason (R): Magnitude of strains in the perpendicular directions of applied stress is smaller than that in the direction of applied stress. [IAS2004] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS21. Ans. (b) IAS22.
Assertion (A): A plane state of stress will, in general, not result in a plane state of strain. [IAS2002] Reason (R): A thin plane lamina stretched in its own plane will result in a state of plane strain. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS22. Ans. (c) R is false. Stress in one plane always induce a lateral strain with its orthogonal plane.
Page 102 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
Previous Conventional Questions with Answers Conventional Question IES1999 Question: Answer:
What are principal in planes? The planes which pass through the point in such a manner that the resultant stress across them is totally a normal stress are known as principal planes. No shear stress exists at the principal planes.
Conventional Question IES2009 Q.
The Mohrâ€™s circle for a plane stress is a circle of radius R with its origin at + 2R on V axis. Sketch the Mohrâ€™s circle and determine V max , V min , V av , this situation.
Ans.
Here V max
V min
Wxy max for [2 Marks]
3R
R
3R R 2R 2 V max V min and W xy 2 V Vv
3R R 2
R
R
R
(2R,0)
3R
Conventional Question IES1999 Question:
Direct tensile stresses of 120 MPa and 70 MPa act on a body on mutually perpendicular planes. What is the magnitude of shearing stress that can be applied so that the major principal stress at the point does not exceed 135 MPa? Determine the value of minor principal stress and the maximum shear stress.
Answer:
Let shearing stress is 'U ' MPa. The principal stresses are 2
Â?120 70 ÂŹÂ 120 70 2 o ÂžÂž Â U ÂžÂ&#x; ÂŽÂ 2 2 Major principal stress is Äą1,2
120 70 Äą1
2
2
Â?120 70 ÂŹÂ 2 ÂÂ U ÂžÂžÂž Â&#x; ÂŽ 2
135(Given) or , U 31.2MPa. Page 103 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
Minor principal stress is Äą2
W max
120 70 2 V1 V 2 2
2
Â§ 120 70 Âˇ 2 Â¨ Â¸ 31.2 2 ÂŠ Âš 135 55 40 MPa 2
55 MPa
Conventional Question IES2009 Q.
The state of stress at a point in a loaded machine member is given by the principle stresses. [ 2 Marks] V1 600 MPa, V 2 0 and V3 600 MPa . (i) (ii)
Ans.
What is the magnitude of the maximum shear stress? What is the inclination of the plane on which the maximum shear stress acts with respect to the plane on which the maximum principle stress V1 acts? (i) Maximum shear stress,
W
V1 V3 2
600 600
2 600 MPa
Ç‰ = 45Âş max. shear stress occurs with Ç”1 plane. Since Ç”1 and Ç”3 are principle stress does not contains shear stress. Hence max. shear stress is at 45Âş with principle (ii) At plane.
Conventional Question IES2008 Question:
Answer:
A prismatic bar in compression has a cross sectional area A = 900 mm2 and carries an axial load P = 90 kN. What are the stresses acts on (i) A plane transverse to the loading axis; (ii) A plane at Č™ = 60o to the loading axis? (i) From figure it is clear A plane transverse to loading axis, Č™ =0o P 90000 N / mm 2 = Tn cos2 Č™= A 900 100 N / mm 2 P 90000 and U = Sin 2Č™= q sinČ™=0 2A 2Ă—900 (iii)
A plane at 60o to loading axis, Č™ = 60Â° 30Â° = 30Â° P 90000 Tn cos2 Č™= q cos2 30 A 900 75N / mm 2 P 90000 U sin2R sin2 q 60o 2A 2 q 900 43.3N / mm 2
Conventional Question IES2001 Question:
A tension member with a crosssectional area of 30 mm2 resists a load of 80 kN, Calculate the normal and shear stresses on the plane of maximum shear stress.
Answer:
Äąn
P cos2 Č™ A
U
P sin 2Č™ 2A
Page 104 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
For maximum shear stress sin2Č™ = 1, or, Č™ = 45o
Äą n
80 q103 P 80 q103 q cos2 45 1333MPa and Umax 1333MPa 30 2A 30 q 2
Conventional Question IES2007 Question:
Answer:
At a point in a loaded structure, a pure shear stress state U = o 400 MPa prevails on two given planes at right angles. (i) What would be the state of stress across the planes of an element taken at +45Â° to the given planes? (ii) What are the magnitudes of these stresses? (i) For pure shear
Äą x Äą y ;
U max oÄą x o400MPa
(ii) Magnitude of these stresses
Äą n U xy Sin 2Č™ U xy Sin90o U xy 400 MPa and U (U xy cos 2Č™ ) 0 Conventional Question IAS1997 Question:
Answer:
Draw Mohr's circle for a 2dimensional stress field subjected to (a) Pure shear (b) Pure biaxial tension (c) Pure uniaxial tension and (d) Pure uniaxial compression Mohr's circles for 2dimensional stress field subjected to pure shear, pure biaxial tension, pure uniaxial compression and pure uniaxial tension are shown in figure below:
Conventional Question IES2003 Question:
A Solid phosphor bronze shaft Page60 105mm of 429in diameter is rotating at 800 rpm and transmitting power. It is subjected torsion only. An electrical resistance
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
strain gauge mounted on the surface of the shaft with its axis at 45Â° to the shaft axis, gives the strain reading as 3.98 Ă— 10â€“4. If the modulus of elasticity for bronze is 105 GN/m2 and Poisson's ratio is 0.3, find the power being transmitted by the shaft. Bending effect may be neglected. Answer:
Let us assume maximum shear stress on the crosssectional plane MU is U . Then
1 4U 2 = U (compressive) 2 1 Principal stress along, LU = 4U 2 U (tensile) 2 Thus magntude of the compressive strain along VM is Principal stress along, VM = 
U = (1 Îź) 3.98 q104 E 3.98 q104 q 105 q109
or U = 32.15 MPa 1 0.3
= Torque being transmitted (T) = U q 32.15 q106 q
ĘŒ qd 3 16
ĘŒ q 0.063 =1363.5 Nm 16
Â? 2ĘŒN ÂŹÂ Â? 2ĘŒĂ—800 ÂŹÂ ÂžÂž = Power being transmitted, P =T.X =T.ÂžÂž =1363.5Ă— Â ÂW 114.23 kW Â&#x;Âž 60 ÂŽÂ Â&#x;Âž 60 ÂŽÂ
Conventional Question IES2002 Question:
The magnitude of normal stress on two mutually perpendicular planes, at a point in an elastic body are 60 MPa (compressive) and 80 MPa (tensile) respectively. Find the magnitudes of shearing stresses on these planes if the magnitude of one of the principal stresses is 100 MPa (tensile). Find also the magnitude of the other principal stress at this point.
Page 106 of 429
Chapter2 Answer:
Principal Stress and Strain
S K Mondalâ€™s
Above figure shows stress condition assuming shear stress is ' U xy'
80Mpa
Jxy
Principal stresses
Äą1,2
2
Â? Äą Äą y ÂŹÂ Â U xy2 o ÂžÂž x ÂžÂ&#x; 2 ÂŽÂÂ
Äąx Äą y 2
or , Äą1,2
60Mpa
60Mpa
Jxy
Jxy
2
Â? 60 80 ÂŹÂ 2 ÂžÂž ÂÂ U xy ÂžÂ&#x; ÂŽ 2
60 80 o 2
2
Jxy 80Mpa
Â? 60 80 ÂŹÂ 60 80 2 or , Äą1,2 o ÂžÂž ÂÂ U xy Âž Â&#x; ÂŽ 2 2 To make principal stress 100 MPa we have to consider '+' . = Äą1 100 MPa 10 702 U xy2 ; or, U xy 56.57 MPa Therefore other principal stress will be 60 80 Äą2 2
2
Â? 60 80 ÂŹÂ 2 Â (56.57) ÂžÂžÂ&#x;Âž ÂŽÂ 2
i.e. 80 MPa(compressive) Conventional Question IES2001 Question: A steel tube of inner diameter 100 mm and wall thickness 5 mm is subjected to a torsional moment of 1000 Nm. Calculate the principal stresses and orientations of the principal planes on the outer surface of the tube. Answer:
Polar moment of Inertia (J)=
ĘŒ Â 4 4 Â˘ 0.110 0.100 ÂŻÂ°Âą = 4.56 q106 m 4 ÂĄ 32
5mm
T .R 1000 q(0.055) T U Now or J J R J 4.56 q106 12.07MPa Now, tan 2Č™ p
2U xy
r,
Äąx Äąy
0
gives Č™p 45 or 135
50mm
0
= Äą1 U xy Sin 2Č™ 12.07 q sin 900 12.07 MPa and Äą 2 12.07 sin 2700 12.07MPa Conventional Question IES2000 Question:
At a point in a two dimensional stress system the normal stresses on two mutually perpendicular planes are T x and T y and the shear stress is U xy. At what value of shear stress, one of the principal stresses will become zero?
Answer:
Two principal stressdes are Äą1,2
Äąx Äą y 2
2
Â? Äą x  Äą y ÂŹÂ Â U xy2 o ÂžÂž ÂžÂ&#x; 2 ÂŽÂÂ Page 107 of 429
Chapter2
Principal Stress and Strain
S K Mondalâ€™s
Considering ()ive sign it may be zero 2
2
Â? Äą x Äą y ÂŹÂ Â? Äą Äą y ÂŹÂ Â U xy2 = ÂžÂž x ÂÂ ÂžÂž ÂžÂ&#x; 2 ÂŽÂÂ ÂžÂ&#x; 2 ÂŽÂ 2
2
Â? Äą Äą y ÂŹÂ Â? Äą x Äą y ÂŹÂ Â Âž Â U xy2 or, ÂžÂž x ÂžÂ&#x; 2 ÂŽÂÂ Â&#x;ÂžÂž 2 ÂŽÂÂ
2
Â? Äą Äą y ÂŹÂ Â? Äą x Äą y ÂŹÂ or, U ÂžÂž x Â ÂžÂž Â Â&#x;Âž 2 ÂŽÂÂ Â&#x;Âž 2 ÂŽÂÂ 2 xy
2 or, U xy Äą xÄą y
or,U xy o Äą x Äą y
Conventional Question IES1996 Question:
Answer:
A solid shaft of diameter 30 mm is fixed at one end. It is subject to a tensile force of 10 kN and a torque of 60 Nm. At a point on the surface of the shaft, determine the principle stresses and the maximum shear stress. Given: D = 30 mm = 0.03 m; P = 10 kN; T= 60 Nm
Pr incipal stresses V 1,V 2 and max imum shear stress W max : Tensile stress V t
10 u 103
Vx
S 4
u 0.032
T J TR J
As per torsion equation,
W
? Shear stress,
14.15 u 106 N / m2 or 14.15 MN / m2
W R 60 u 0.015
TR
S 32
D
S
4
32
u 0.03
4
11.32 u 106 N / m2
or 11.32 MN / m2 The principal stresses are calculated by u sin g the relations : Â§Vx Vy Âˇ Â¸r 2 Âš ÂŠ
V 1,2 Â¨ Here
Vx
ÂŞÂ§ V x V y Âˇ 2 Âş 2 ÂŤÂ¨ Â¸ Âť W xy 2 Âš Âť ÂŤÂŹÂŠ Âź
14.15MN / m2 ,V y
0;W xy
W
11.32
MN / m2
2
?
V 1,2
14.15 2 Â§ 14.15 Âˇ r Â¨ 11.32
Â¸ 2 ÂŠ 2 Âš 7.07 r 13.35
Hence,major principal stress, V 1 Minor principal stress, V 2 Maximum shear stress,W max
20.425 MN / m2 , 6.275MN / m2 . 20.425 MN / m2 tensile
6.275MN / m2 compressive
V1 V 2
24.425 6.275
2
2
13.35mm / m2
Conventional Question IES2000 Two planes AB and BC which are at right angles are acted upon by tensile stress of 140 N/mm2 and a compressive stress of 70 N/mm2 respectively and also by stress 35 N/mm2. Determine the principal stresses and principal planes. Find also the maximum shear stress and planes on which they act. Sketch the Mohr circle and mark the relevant data.
Question:
Page 108 of 429
Chapter2 Answer:
Principal Stress and Strain Given
S K Mondalâ€™s 2
70N/mm
Äą x =140MPa(tensile)
B
C
Äą y = 70MPa(compressive)
35Nmm
U xy 35MPa
2 2
140N/mm
Principal stresses; Äą1, Äą2 ; A
We know that, Äą1,2
Äąx Äąy 2
2
Â? Äą x Äą y ÂŹÂ 2 Â U xy o ÂžÂžÂž ÂžÂ&#x; 2 ÂŽÂÂ
140 70 o 2
2
Â?140 70 ÂŹÂ 2 ÂžÂž ÂÂ 35 35 o 110.7 ÂžÂ&#x; ÂŽ 2
Therefore Äą1 =145.7 MPa and Äą 2 75.7MPa
Position of Principal planes Č™1, Č™ 2 tan 2Č™ p
2U xy Äąx Äąy
2 q 35 0.3333 140 70
Maximum shear stress, Umax
Äą1  Äą 2 145 75.7 110.7MPa 2 2 Y
Mohr cirle: OL=Äą x 140MPa
S
OM Äą y 70MPa SM LT U xy 35MPa
U
2Rp =198.4
L M
O
Joining ST that cuts at 'N'
N
V
2R
Äą = 140
T
SN=NT=radius of Mohr circle =110.7 MPa OV=Äą1 145.7MPa OV Äą 2 75.7MPa
Conventional Question IES2010 Q6.
The data obtained from a rectangular strain gauge rosette attached to a stressed
steel
member
0 Â?0 = 220 u 106 , Â?45 120 u 1006 ,
are
and
Â?90 =220 u 106 . Given that the value of E = 2 u 105 N / mm2 and Poissonâ€™s Ratio P 0.3 , calculate the values of principal stresses acting at the point and Ans.
their directions. A rectangular strain gauge rosette strain
Â?0 220 u 106 E = 2 u 1011 N / m2
Â?
450
[10 Marks]
120 u 106 Â?90 220 u 106 poisson ratio P 0.3
Find out principal stress and their direction. Let ea Â?o ec Â?90 and eb Â?45 We know that principal strain are
ea eb 2 2 ea eb eb ec
2 220 u 106 120 u 106 1 Â&#x; r 220 120 106 2 2 1 354.40 u 106 Â&#x; 50 u 106 r 2 Page 109 of 429 Â?12
2
120 220 10
6
2
Chapter2 6
Â?12 Â&#x; 50 u 10
Principal Stress and Strain r 250.6 u 106
S K Mondalâ€™s
Â?1 2.01 u 104
Â?2 3.01 u 104 Direction can be find out : 
2 u 120 u 106 220 u 106 220 u 106
2eb ea ec ec ea
tan 2Ç‰pÂ?
240 0.55 440 2Ç‰pÂ? 28.81 Â&#x;
Ç‰pÂ?
14.450 clockwiseform principal strain t1
Principal stress:
Ç”1
E Â?1 Ç? Â?2
2 u 1011 2 0.3 3 u 104
2
2
1 Ç?
1 0.3
241.78 u 105 N / m 2 527.47 u 105 N / m 2 Conventional Question IES1998 Question: Answer:
When using straingauge system for stress/force/displacement measurements how are inbuilt magnification and temperature compensation achieved? Inbuilt magnification and temperature compensation are achieved by (a) Through use of adjacent arm balancing of Wheatstone bridge. (b) By means of self temperature compensation by selected meltgauge and dual elementgauge.
Conventional Question AMIE1998 Question:
Answer:
A cylinder (500 mm internal diameter and 20 mm wall thickness) with closed ends is subjected simultaneously to an internal pressure of 060 MPa, bending moment 64000 Nm and torque 16000 Nm. Determine the maximum tensile stress and shearing stress in the wall. Given: d = 500 mm = 0Âˇ5 m; t = 20 mm = 0Âˇ02 m; p = 0Âˇ60 MPa = 0.6 MN/m2; M = 64000 Nm = 0Âˇ064 MNm; T= 16000 Nm = 0Âˇ016 MNm. Maximum tensile stress: First let us determine the principle stresses V 1 and V 2 assuming this as a thin cylinder.
and
V2
pd 0.6 u 0.5 7.5MN / m2 2t 2 u 0.02 0.6 u 0.5 3.75MN / m2 4 u 0.02
V1
We know,
pd 4t
Next consider effect of combined bending moment and torque on the walls of the cylinder. Then the principal stresses V '1 and V '2 are given by
Page 110 of 429
Chapter2
Principal Stress and Strain
V '1 and
V '2
?
V '1
and
V '2
16 Âª M M2 T 2 Âº 3 Â¬ Â¼ Sd 16 Âª M M2 T 2 Âº Â¼ S d3 Â¬ 16 Âª0.064 0.0642 0.0162 Âº 3 Â¬ Â¼ S u 0.5
16
S u 0.5
3
Âª0.064 0.0642 0.0162 Âº Â¬ Â¼
S K Mondalâ€™s
5.29MN / m2
0.08MN / m2
Maximum shearing stress,W max : We Know, W max
V II ?
W max
V I V II 2
V 2 V '2
3.75 0.08
12.79 3.67 2
3.67MN / m2 tensile
4.56MN / m2
Page 111 of 429
3.
Moment of Inertia and Centroid
Theory at a Glance (for IES, GATE, PSU) 3.1 Centre of gravity The centre of gravity of a body defined as the point through which the whole weight of a body may be assumed to act.
3.2 Centroid or Centre of area The centroid or centre of area is defined as the point where the whole area of the figure is assumed to be concentrated.
3.3 Moment of Inertia (MOI) x
About any point the product of the force and the perpendicular distance between them is known as moment of a force or first moment of force.
x
This first moment is again multiplied by the perpendicular distance between them to obtain second moment of force.
x
In the same way if we consider the area of the figure it is called second moment of area or area moment of inertia and if we consider the mass of a body it is called second moment of mass or mass moment of Inertia.
x
Mass moment of inertia is the measure of resistance of the body to rotation and forms the basis of dynamics of rigid bodies.
x
Area moment of Inertia is the measure of resistance to bending and forms the basis of strength of materials.
3.4 Mass moment of Inertia (MOI)
I
¦m r
2 i i
i
x
Notice that the moment of inertia ‘I’ depends on the distribution of mass in the system.
x
The furthest the mass is from the rotation axis, the bigger the moment of inertia.
x
For a given object, the moment of inertia depends on where we choose the rotation axis.
x
In rotational dynamics, the moment of inertia ‘I’ appears in the same way that mass m does Page 112 of 429
in linear dynamics.
Chapter3 x
Moment of Inertia and Centroid
Solid disc or cylinder of mass M and radius R, about perpendicular axis through its
1 MR2 2
centre, I
x
Solid sphere of mass M and radius R, about an axis through its centre, I = 2/5 M R2
x
Thin rod of mass M and length L, about a perpendicular axis through its centre.
I x
1 ML2 12
Thin rod of mass M and length L, about a perpendicular axis through its end.
I
1 ML2 3
3.5 Area Moment of Inertia (MOI) or Second moment of area x
To find the centroid of an area by the first moment of the area about an axis was determined ( 혵 x dA )
x
Integral of the second moment of area is called moment of inertia (혵 x2dA)
x
Consider the area ( A )
x
By definition, the moment of inertia of the differential area about the x and y axes are dIxx and dIyy
x
dIxx = y2dA
Ixx = 혵 y2 dA
x
dIyy = x2dA
Iyy = 혵 x2 dA
3.6 Parallel axis theorem for an area The rotational inertia about any axis is the sum of second moment of inertia about a parallel axis through the C.G and total area of the body times square of the distance between the axes.
INN = ICG + Ah2
Page 113 of 429
Chapter3
Moment of Inertia and Centroid
3.7 Perpendicular axis theorem for an area If x, y & z are mutually perpendicular axes as shown, then I zz J
I xx I yy
Zaxis is perpendicular to the plane of x â€“ y and vertical to this page as shown in figure.
x
To find the moment of inertia of the differential area about the pole (point of origin) or zaxis,
(r) is used. (r) is the perpendicular distance from the pole to dA for the entire area J = Âœ r2 dA = Âœ (x2 + y2 )dA = Ixx + Iyy (since r2 = x2 + y2 ) Where, J = polar moment of inertia
3.8 Moments of Inertia (area) of some common area (i) MOI of Rectangular area Moment of inertia about axis XX which passes through centroid. Take an element of width â€˜dyâ€™ at a distance y from XX axis.
? Area of the element (dA) = b u dy. and Moment of Inertia of the element about XX axis dA u y 2
?Total
b.y 2 .dy
MOI about XX axis (Note it is area
moment of Inertia) h
I xx
2
Âł
h
h
2
by dy 2
2
2 Âł by 2 dy 0
bh3 12
3
bh 12
I xx Similarly, we may find, I yy
hb3 12
?Polar moment of inertia (J) = Ixx + Iyy =
bh3 hb3 12 12
Page 114 of 429
Chapter3
Moment of Inertia and Centroid
If we want to know the MOI about an axis NN passing through the bottom edge or top edge. Axis XX and NN are parallel and at a distance h/2. Therefore INN = Ixx + Area
u (distance) 2
bh3 Â§hÂˇ buhuÂ¨ Â¸ 12 ÂŠ2Âš
2
bh3 3
CaseI: Square area
I xx
a4 12
CaseII: Square area with diagonal as axis
a4 12
I xx
CaseIII:
Rectangular area with a centrally
rectangular hole Moment of inertia of the area = moment of inertia of BIG rectangle â€“ moment of inertia of SMALL rectangle
I xx
BH 3 bh3 12 12
Page 115 of 429
Chapter3 Moment of Inertia and Centroid (ii) MOI of a Circular area The moment of inertia about axis XX this passes through the centroid. It is very easy to find polar moment of inertia about point ‘O’. Take an element of width ‘dr’ at a distance ‘r’ from centre. Therefore, the moment of inertia of this element about polar axis
d(J) = d(Ixx + Iyy ) = area of ring u (radius)2 2S rdr u r 2
or d(J)
Integrating both side we get R
J
S R4
3
³ 2S r dr
S D4
2 32 0 Due to summetry I xx I yy Therefore, I xx
I xx
J 2
I yy
64
S D4
I yy
CaseI:
S D4
64
and J
Moment of inertia of a circular
area with a concentric hole. Moment of inertia of the area = moment of inertia of BIG circle – moment of inertia of SMALL circle.
Ixx = Iyy =
S D4 64
S 64
CaseII:
S d4 64
( D4 d4 )
S
and J
–
32
( D4 d4 )
Moment of inertia of a semi
circular area. I NN
1 of the momemt of total circular lamina 2 1 § S D4 · S D4 u¨ ¸ 2 © 64 ¹ 128
We
know
4r 3S
2D 3S
that
distance
of
CG
from
base
is
h say
i.e. distance of parallel axis XX and NN is (h)
? According to parallel axis theory
Page 116 of 429
S D4 32
Chapter3
Moment of Inertia and Centroid I G Area Ã— distance
I NN or or
S D4
1 Â§ S D2 Â· 2 I xx Â¨ Â¸ u h
2Â© 4 Â¹ 1 Â§ S D 2 Â· Â§ 2D Â· I xx u Â¨ Â¸u 2 Â© 4 Â¹ Â¨Â© 3S Â¸Â¹
128
S D4 128
0.11R 4
I xx
or
2
Case â€“ III: Quarter circle area IXX = one half of the moment of Inertia of the Semicircular area about XX.
1 u 0.11R4 2
I XX
I XX
0.055 R4
0.055 R 4
INN = one half of the moment of Inertia of the Semicircular area about NN.
? I NN
1 S D4 u 2 64
S D4 128
(iii) Moment of Inertia of a Triangular area (a) Moment of Inertia of a Triangular area of a axis XX parallel to base and passes through C.G.
I XX
bh3 36
(b) Moment of inertia of a triangle about an axis passes through base
I NN
bh3 12
Page 117 of 429
Chapter3 Moment of Inertia and Centroid (iv) Moment of inertia of a thin circular ring: Polar moment of Inertia
J
R2 u area of whole ring R 2 u 2S Rt 2S R 3 t
I XX
IYY
J 2
S R 3t
(v) Moment of inertia of a elliptical area
I XX
S ab3 4
Let us take an example: An Isection beam of 100 mm wide, 150 mm depth flange and web of thickness 20 mm is used in a structure of length 5 m. Determine the Moment of Inertia (of area) of crosssection of the beam. Answer: Carefully observe the figure below. It has sections with symmetry about the neutral axis.
We may use standard value for a rectangle about an axis passes through centroid. i.e. I
bh3 . 12
The section can thus be divided into convenient rectangles for each of which the neutral axis passes I Beam
the centroid.
I Re c tan gle  I Shaded area ª 0.100 u 0.150 3 0.40 u 0.1303 º 4 « »m 2 u 12 12 «¬ »¼ 1.183 u 104 m4
3.9 Radius of gyration Consider area A with moment of inertia Ixx. Imagine that the area is concentrated in a thin strip parallel to Page 118 of 429
Chapter3
Moment of Inertia and Centroid
the x axis with equivalent Ixx.
kxx
2 kxx A or
I xx
I xx A
kxx =radius of gyration with respect to the x axis.
Similarly
I yy
2 kyy A or
ko2 A or
J
ko2
I yy
kyy
ko
A
J A
2 2 kxx kyy
Let us take an example: Find radius of gyration for a circular area of diameter â€˜dâ€™ about central axis.
Answer:
We know that, I xx
2 K xx A
Page 119 of 429
Chapter3
Moment of Inertia and Centroid Sd
or K XX
I XX A
4
64 S d2 4
d 4
Page 120 of 429
Chapter3
Moment of Inertia and Centroid
OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20Years GATE Questions Moment of Inertia (Second moment of an area) GATE1.
The second moment of a circular area about the diameter is given by (D is the diameter) (a)
[GATE2003]
S D4
(b)
4
S D4 16
(c)
S D4 32
(d)
S D4 64
GATE1. Ans. (d)
GATE2.
The area moment of inertia of a square of size 1 unit about its diagonal is: [GATE2001] (a)
1 3
GATE2. Ans. (c) I xx
(b)
a4 12
1
1 4
(c)
1 12
(d)
1 6
4
12
Radius of Gyration Data for Q3–Q4 are given below. Solve the problems and choose correct answers. A reel of mass “m” and radius of gyration “k” is rolling down smoothly from rest with one end of the thread wound on it held in the ceiling as depicted in the figure. Consider the thickness of the thread and its mass negligible in comparison with the radius “r” of the hub and the reel mass “m”. Symbol “g” represents the acceleration due to gravity. [GATE2003]
Page 121 of 429
Chapter3
GATE3.
Moment of Inertia and Centroid
The linear acceleration of the reel is: gr 2 gk2 (a) 2 (b) r k2 r 2 k2
(c)
grk
r
2
k2
(d)
mgr 2
r
2
k2
GATE3. Ans. (a) For downward linear motion mg â€“ T = mf, where f = linear tangential acceleration = rÇ‚, Ç‚ = rotational acceleration. Considering rotational motion Tr I D . or, T = mk2 u
GATE4.
f gr 2 therefore mg â€“ T = mf gives f = r2 r 2 k2
The tension in the thread is: mgrk mgr 2 (a) 2 (b) 2 2 r k2 r k
GATE4. Ans. (c) T
mk2 u
f r2
mk2 u
gr 2
(c)
2
2
mgk
r r k
2
r
2
mgk2
r
2
k
2
(d)
mg
r
2
k2
2
k2
Previous 20Years IES Questions Centroid IES1.
Assertion (A): Inertia force always acts through the centroid of the body and is directed opposite to the acceleration of the centroid. [IES2001] Reason (R): It has always a tendency to retard the motion. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES1. Ans. (c) It has always a tendency to oppose the motion not retard. If we want to retard a motion then it will wand to accelerate.
Page 122 of 429
Chapter3
Moment of Inertia and Centroid
Radius of Gyration IES2.
Figure shows a rigid body of mass m having radius of gyration k about its centre of gravity. It is to be replaced by an equivalent dynamical system of two masses placed at A and B. The mass at A should be: a um bum (a) (b) a b a b m a m b (d) (c) u u 3 b 2 a [IES2003]
IES2. Ans. (b) IES3. Force required to accelerate a cylindrical body which rolls without slipping on a horizontal plane (mass of cylindrical body is m, radius of the cylindrical surface in contact with plane is r, radius of gyration of body is k and acceleration of the body is a) is: [IES2001] 2 2 2 2 2 2 (b) mk / r .a (c) mk .a (a) m k / r 1 .a (d) mk / r 1 .a
IES3. Ans. (a) IES4. A body of mass m and radius of gyration k is to be replaced by two masses m1 and m2 located at distances h1 and h2 from the CG of the original body. An equivalent dynamic system will result, if [IES2001] (a) h1 h2
k
2
(b) h1
h22
k2
(c) h1h2
k2
(d)
h1h2
k2
IES4. Ans. (c)
Previous 20Years IAS Questions Radius of Gyration IAS1.
A wheel of centroidal radius of gyration 'k' is rolling on a horizontal surface with constant velocity. It comes across an obstruction of height 'h' Because of its rolling speed, it just overcomes the obstruction. To determine v, one should use the principle (s) of conservation of [IAS 1994] (a) Energy (b) Linear momentum (c) Energy and linear momentum (d) Energy and angular momentum IAS1. Ans. (a)
Page 123 of 429
Chapter3
Moment of Inertia and Centroid
Previous Conventional Questions with Answers Conventional Question IES2004 Question: Answer:
When are Isections preferred in engineering applications? Elaborate your answer. Isection has large section modulus. It will reduce the stresses induced in the material. Since Isection has the considerable area are far away from the natural so its section modulus increased.
Page 124 of 429
4.
Bending Moment and Shear Force Diagram
Theory at a Glance (for IES, GATE, PSU) 4.1 Shear Force and Bending Moment At first we try to understand what shear force is and what is bending moment? We will not introduce any other coordinate system. We use general coordinate axis as shown in the figure. This system will be followed in shear force and bending moment diagram and in deflection of beam. Here downward direction will be negative i.e. negative Yaxis. Therefore downward deflection of the
We use above Coordinate system
beam will be treated as negative.
Some books fix a coordinate axis as shown in the following figure. Here downward direction will be positive i.e. positive Yaxis. Therefore downward deflection of the beam will be treated as positive. As beam is generally deflected in downward directions and
this
coordinate
system
treats
downward
deflection is positive deflection. Consider a cantilever beam as shown subjected to external load ‘P’. If we imagine this beam to be cut by a section XX, we see that the applied force tend to displace the lefthand portion of the beam relative to the right hand portion, which is fixed in the wall. This tendency is resisted by internal forces between the two parts of the beam. At the cut section a resistance shear force (Vx) and a bending moment (Mx) is induced. This resistance shear force and the bending moment at the cut section is shown in the left hand and right hand portion of the cut beam. Using the three equations of equilibrium
¦F
x
0 ,
¦F
y
We find that Vx
0 and
¦M
P and M x
i
0
P . x
In this chapter we want to show pictorially Page 125the of 429
Some books use above coordinate system
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
variation of shear force and bending moment in a beam as a function of ‘x' measured from one end of the beam.
Shear Force (V) ŋ equal in magnitude but opposite in direction to the algebraic sum (resultant) of the components in the direction perpendicular to the axis of the beam of all external loads and support reactions acting on either side of the section being considered. Bending Moment (M) equal in magnitude but opposite in direction to the algebraic sum of the moments about (the centroid of the cross section of the beam) the section of all external loads and support reactions acting on either side of the section being considered.
What are the benefits of drawing shear force and bending moment diagram? The benefits of drawing a variation of shear force and bending moment in a beam as a function of ‘x' measured from one end of the beam is that it becomes easier to determine the maximum absolute value of shear force and bending moment. The shear force and bending moment diagram gives a clear picture in our mind about the variation of SF and BM throughout the entire section of the beam. Further, the determination of value of bending moment as a function of ‘x' becomes very important so as to determine the value of deflection of beam subjected to a given loading where we will use the formula, EI
d 2y dx 2
Mx .
4.2 Notation and sign convention x Shear force (V) Positive Shear Force A shearing force having a downward direction to the right hand side of a section or upwards to the left hand of the section will be taken as ‘positive’. It is the usual sign conventions to be followed for the shear force. In some book followed totally opposite sign convention.
Page 126 of 429
Chapter4
Bending Moment and Shear Force Diagram
downward
S K Mondal’s
The upward direction shearing
The
direction
force which is on the left hand
shearing force which is on the
of the section XX is positive
right hand of the section XX is
shear force.
positive shear force.
Negative Shear Force A shearing force having an upward direction to the right hand side of a section or downwards to the left hand of the section will be taken as ‘negative’.
The
x
downward
direction
The upward direction shearing
shearing force which is on the
force which is on the right
left hand of the section XX is
hand of the section XX is
negative shear force.
negative shear force.
Bending Moment (M) Positive Bending Moment A bending moment causing concavity upwards will be taken as ‘positive’ and called as sagging bending moment.
Page 127 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
Sagging If the bending moment of
If the bending moment of
A bending moment causing
the left hand of the section
the right hand of the
concavity upwards will be
XX is clockwise then it is a
section
positive bending moment.
clockwise then it is a
called as sagging bending
positive bending moment.
moment.
XX
is
anti
taken
as
‘positive’
and
Negative Bending Moment
Hogging If the bending moment of
A bending moment causing
the
the right hand of the
convexity upwards will be
anti
section XX is clockwise
taken as ‘negative’ and called
If the bending moment of the
left
section
hand XX
is
of
clockwise then it is a
then
positive bending moment.
bending moment.
it
is
a
positive
as hogging bending moment.
Way to remember sign convention
x
Remember in the Cantilever beam both Shear force and BM are negative (–ive).
4.3 Relation between S.F (Vx), B.M. (Mx) & Load (w) x
dVx = w (load) The value of the distributed load at any point in the beam is dx equal to the slope of the shear force curve. (Note that the sign of this rule may change depending on the sign convention used for the external distributed load).
x
dM x = Vx The value of the shear force at any point in the beam is equal to the slope dx of the bending moment curve.
Page 128 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
4.4 Procedure for drawing shear force and bending moment diagram Construction of shear force diagram x
From the loading diagram of the beam constructed shear force diagram.
x
First determine the reactions.
x
Then the vertical components of forces and reactions are successively summed from the left end of the beam to preserve the mathematical sign conventions adopted. The shear at a section is simply equal to the sum of all the vertical forces to the left of the section.
x
The shear force curve is continuous unless there is a point force on the beam. The curve then “jumps” by the magnitude of the point force (+ for upward force).
x
When the successive summation process is used, the shear force diagram should end up with the previously calculated shear (reaction at right end of the beam). No shear force acts through the beam just beyond the last vertical force or reaction. If the shear force diagram closes in this fashion, then it gives an important check on mathematical calculations. i.e. The shear force will be zero at each end of the beam unless a point force is applied at the end.
Construction of bending moment diagram
x
The bending moment diagram is obtained by proceeding continuously along the length of beam from the left hand end and summing up the areas of shear force diagrams using proper sign convention.
x
The process of obtaining the moment diagram from the shear force diagram by summation is exactly the same as that for drawing shear force diagram from load diagram.
x
The bending moment curve is continuous unless there is a point moment on the beam. The curve then “jumps” by the magnitude of the point moment (+ for CW moment).
x
We know that a constant shear force produces a uniform change in the bending moment, resulting in straight line in the moment diagram. If no shear force exists along a certain portion of a beam, then it indicates that there is no change in moment takes place. We also know that dM/dx= Vx therefore, from the fundamental theorem of calculus the maximum or minimum moment occurs where the shear is zero.
x
The bending moment will be zero at each free or pinned end of the beam. If the end is built in, the moment computed by the summation must be equal to the one calculated initially for the reaction.
4.5 Different types of Loading and their S.F & B.M Diagram (i) A Cantilever beam with a concentrated load ‘P’ at its free end.
Page 129 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
Shear force: At a section a distance x from free end consider the forces to the left, then (Vx) =  P (for all values of x) negative in sign i.e. the shear force to the left of the xsection are in downward direction and therefore negative.
Bending Moment: Taking moments about the section gives (obviously to the left of the section)
Mx = P.x (negative sign means that the
S.F and B.M diagram
moment on the left hand side of the portion is in the anticlockwise direction and is therefore taken as negative according to the sign convention) so that the maximum bending moment occurs at the fixed end i.e. Mmax =  PL (at x = L)
(ii) A Cantilever beam with uniformly distributed load over the whole length When a cantilever beam is subjected to a uniformly distributed load whose intensity is given w /unit length.
Shear force: Consider any crosssection XX which is at a distance of x from the free end. If we just take the resultant of all the forces on the left of the Xsection, then
Vx = w.x
for all values of â€˜x'.
At x = 0, Vx = 0 At x = L, Vx = wL (i.e. Maximum at fixed end) Plotting the equation Vx = w.x, we get a straight line because it is a equation of a straight line y (Vx) = m( w) .x
Bending Moment: Bending Moment at XX is obtained by treating the load to the left of XX as a concentrated load of the same value (w.x)
S.F and B.M diagram
acting through the centre of gravity at x/2. Therefore, the bending moment at any crosssection XX is
Mx
w .x .
x 2
w .x 2 2
Therefore the variation of bending moment is according to parabolic law. The extreme values of B.M would be at x = 0,
Mx = 0
and x = L, Mx =
wL2 2
Page 130 of 429
Chapter4
Bending Moment and Shear Force Diagram
Maximum bending moment,
Mmax
wL2 2
S K Mondalâ€™s
at fixed end
Another way to describe a cantilever beam with uniformly distributed load (UDL) over itâ€™s whole length.
(iii) A Cantilever beam loaded as shown below draw its S.F and B.M diagram
In the region 0 < x < a Following the same rule as followed previously, we get
Vx = P; and Mx =  P.x In the region a < x < L
Vx = P +P=0; and Mx =  P.x +P x a P.a
S.F and B.M diagram
(iv) Let us take an example: Consider a cantilever bean of 5 m length. It carries a uniformly distributed load 3 KN/m and a concentrated load of 7 kN at the free end and 10 kN at 3 meters from the fixed end.
Draw SF and BM diagram.
Page 131 of 429
Chapter4
Bending Moment and Shear Force Diagram
Answer: In the region 0 < x < 2 m Consider any cross section XX at a distance x from free end. Shear force (Vx) = 7 3x So, the variation of shear force is linear. at x = 0,
Vx = 7 kN
at x = 2 m , Vx = 7  3 u 2 = 13 kN at point Z
Vx = 7 3 u 210 = 23 kN
Bending moment (Mx) = 7x  (3x).
x 2
3x 2 7x 2
So, the variation of bending force is parabolic. at x = 0,
Mx = 0
at x = 2 m,
Mx = 7 u 2 â€“ (3 u 2) u
2 =  20 kNm 2
In the region 2 m < x < 5 m Consider any cross section YY at a distance x from free end Shear force (Vx) = 7  3x â€“ 10 = 17 3x So, the variation of shear force is linear. at x = 2 m, Vx =  23 kN at x = 5 m, Vx =  32 kN
Â§xÂˇ Â¸  10 (x  2) ÂŠ2Âš
Bending moment (Mx) =  7x â€“ (3x) u Â¨
3 x 2 17 x 20 2 So, the variation of bending force is parabolic. at x = 2 m, Mx
3 u 22 17 u 2 20 =  20 kNm 2
at x = 5 m, Mx =  102.5 kNm
Page 132 of 429
S K Mondalâ€™s
Chapter4
(v)
Bending Moment and Shear Force Diagram
S K Mondal’s
A Cantilever beam carrying uniformly varying load from zero at free end and w/unit length at the fixed end
Consider any crosssection XX which is at a distance of x from the free end. At this point load (wx) =
Therefore total load (W)
w .x L L
L
³ w x dx
³ L .xdx =
0
Shear force Vx
w
0
wL 2
area of ABC (load triangle)
1 §w · wx 2 . ¨ x ¸ .x 2 ©L ¹ 2L ? The shear force variation is parabolic. at x = 0, Vx 0 at x = L, Vx
WL i.e. Maximum Shear force (Vmax ) 2
Page 133 of 429
WL at fixed end 2
Chapter4 Bending Moment and Shear Force Diagram Bending moment Mx load u distance from centroid of triangle ABC
S K Mondal’s
wx 2 § 2x · wx 3 .¨ ¸ 2L © 3 ¹ 6L ? The bending moment variation is cubic. at x= 0, Mx 0
at x = L,
Mx
wL2 i.e. Maximum Bending moment (Mmax ) 6
Integration method
d Vx
w load .x
Alternative way : We know that
dx
L
or d(Vx )
w .x .dx L
Integrating both side Vx
x
³ d Vx
0
or Vx
³ 0
w . x .dx L
w x2 . L 2
Again we know that d Mx
Vx
dx or
d Mx


wx 2 2L
wx 2 dx 2L
Page 134 of 429
wL2 at fixed end. 6
Chapter4 Bending Moment and Shear Force Diagram Integrating both side we get at x=0,Mx =0
Mx
Âł
x
d(Mx )
0
or Mx
Âł 0

S K Mondalâ€™s
wx 2 .dx 2L
w x3 Ă— 2L 3

wx 3 6L
(vi) A Cantilever beam carrying gradually varying load from zero at fixed end and w/unit length at the free end
Considering equilibrium we get, MA
wL2 and Reaction R A
3
wL 2
Considering any crosssection XX which is at a distance of x from the fixed end. At this point load (Wx )
Shear force Vx
W .x L
RA area of triangle ANM
wL 1 Â§ w Âˇ wL wx 2  . Â¨ .x Â¸ .x = + 2 2 ÂŠL Âš 2 2L ? The shear force variation is parabolic. wL at x 0, Vx i.e. Maximum shear force, Vmax 2 at x L, Vx 0 Bending moment Mx =R A .x 
wL 2
wx 2 2x .  MA 2L 3
wL wx 3 wL2 .x 2 6L 3 ? The bending moment variation is cubic =
at x = 0, Mx
at x
0
L, Mx
wL2 i.e.Maximum B.M. Mmax
3
Page 135 of 429
wL2 . 3
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
(vii) A Cantilever beam carrying a moment M at free end
Consider any crosssection XX which is at a distance of x from the free end.
Shear force: Vx = 0 at any point. Bending moment (Mx) = M at any point, i.e. Bending moment is constant throughout the length.
(viii) A Simply supported beam with a concentrated load ‘P’ at its mid span.
Considering equilibrium we get, R A = RB =
P 2
Now consider any crosssection XX which is at a distance of x from left end A and section YY at a distance from left end A, as shown in figure below.
Page 136 of 429
Shear force: In the region 0 < x < L/2
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
Vx = RA = + P/2 (it is constant)
In the region L/2 < x < L Vx = RA – P =
P  P =  P/2 2
(it is constant)
Bending moment: In the region 0 < x < L/2 Mx =
P .x 2
(its variation is linear)
at x = 0, Mx = 0
and at x = L/2 Mx =
Maximum bending moment,
PL i.e. maximum 4
PL 4
Mmax
at x = L/2 (at midpoint)
In the region L/2 < x < L Mx =
PL P P .x (its variation is linear) .x – P(x  L/2) = 2 2 2
at x = L/2 , Mx =
PL 4
and at x = L,
Mx = 0
(ix) A Simply supported beam with a concentrated load ‘P’ is not at its mid span.
Considering equilibrium we get, RA =
Pb Pa and RB = L L
Now consider any crosssection XX which is at a distance x from left end A and another section YY at a distance x from end A as shownPage in figure below. 137 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
Shear force: In the range 0 < x < a Vx = RA = +
Pb L
(it is constant)
In the range a < x < L Vx = RA  P = 
Pa L
(it is constant)
Bending moment: In the range 0 < x < a Mx = +RA.x =
Pb .x L
at x = 0, Mx = 0
(it is variation is linear)
and at x = a, Mx =
Pab L
(i.e. maximum)
In the range a < x < L Mx = RA.x – P(x a) =
§ ©
Pb .x – P.x + Pa (Put b = L  a) L
= Pa (1  Pa ¨ 1 at x = a, Mx =
x· ) L ¸¹
Pab L
and at x = L, Mx = 0
(x) A Simply supported beam with two concentrated load ‘P’ from a distance ‘a’ both end. The loading is shown below diagram
Page 138 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
Take a section at a distance x from the left support. This section is applicable for any value of x just to the left of the applied force P. The shear, remains constant and is +P. The bending moment varies linearly from the support, reaching a maximum of +Pa. A section applicable anywhere between the two applied forces. Shear force is not necessary to maintain equilibrium of a segment in this part of the beam. Only a constant bending moment of +Pa must be resisted by the beam in this zone. Such a state of bending or flexure is called pure bending. Shear and bendingmoment diagrams for this loading condition are shown below.
(xi) A Simply supported beam with a uniformly distributed load (UDL) through out its length
We will solve this problem by following two alternative ways.
(a) By Method of Section Considering equilibrium we get RA = RB =
wL 2
Now Consider any crosssection XX which is at a distance x from left end A.
Page 139 of 429
Chapter4
Bending Moment and Shear Force Diagram
Then the section view
Shear force: Vx =
wL wx 2
(i.e. S.F. variation is linear) Vx =
at x = 0,
wL 2
at x = L/2, Vx = 0 Vx = 
at x = L,
wL 2
wL wx 2 .x 2 2
Bending moment: M x
(i.e. B.M. variation is parabolic) at x = 0, Mx = 0 at x = L, Mx = 0 Now we have to determine maximum bending moment and its position.
For maximum B.M:
d Mx
dx or
0 i .e. Vx
wL wx 2
0 or
Therefore, maximum bending moment,
0
x
ª d M x
« dx ¬
L 2
Mmax
wL2 8
(a) By Method of Integration Shear force: We know that,
or
d Vx
dx
d Vx
w
wdx
Integrating both side we get (at x =0, Vx =
º Vx » ¼
wL ) 2
Page 140 of 429
at x = L/2
S K Mondal’s
Chapter4
Bending Moment and Shear Force Diagram
Vx
S K Mondalâ€™s
x
Â³ d V
Â³ wdx
x
wL 2
or Vx or Vx
0
wL wx 2 wL wx 2
Bending moment: We know that,
d Mx
dx
Vx
d M x Vx dx
or
Â§ wL Â· Â¨ 2 wx Â¸ dx Â© Â¹
Integrating both side we get (at x =0, Vx =0) Mx
Â³
x
d Mx
o
or M x
Â§ wL Â· wx Â¸ dx 2 Â¹ 0
Â³ Â¨Â©
wL wx 2 .x 2 2
Let us take an example: A loaded beam as shown below. Draw its S.F and B.M diagram.
Considering equilibrium we get
Â¦M
A
0 gives
 200 u 4 u 2 3000 u 4 RB u 8 or
RB
1700N
And R A RB or
RA
0
200 u 4 3000
2100N
Now consider any crosssection XX which is at a distance 'x' from left end A and as shown in figure
Page 141 of 429
Chapter4
Bending Moment and Shear Force Diagram
In the region 0 < x < 4m Shear force (Vx) = RA – 200x = 2100 – 200 x
§x· 2 ¸ = 2100 x 100 x ©2¹
Bending moment (Mx) = RA .x – 200 x . ¨ at x = 0, at x = 4m,
Vx = 2100 N, Vx = 1300 N,
Mx = 0 Mx = 6800 N.m
In the region 4 m < x < 8 m Shear force (Vx) = RA  200 u 4 – 3000 = 1700 Bending moment (Mx) = RA. x  200 u 4 (x2) – 3000 (x 4) = 2100 x – 800 x + 1600 – 3000x +12000 = 13600 1700 x at x = 4 m,
Vx = 1700 N,
at x = 8 m,
Vx = 1700 N,
Mx = 6800 Nm Mx = 0
Page 142 of 429
S K Mondal’s
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
(xii) A Simply supported beam with a gradually varying load (GVL) zero at one end and w/unit length at other span.
Consider equilibrium of the beam =
¦M
B
R A .L or R A
1 wL acting at a point C at a distance 2L/3 to the left end A. 2
0 gives wL L . 2 3 wL 6
Similarly
0
¦M
A
0 gives RB
wL 3
The free body diagram of section A  XX as shown below, Load at section XX, (wx) =
Page 143 of 429
w x L
Chapter4
Bending Moment and Shear Force Diagram
The resulted of that part of the distributed load which acts on this free body is applied at a point Z, distance x/3 from XX section.
Shear force (Vx) = R A 
wx 2 2L
wL wx 2 6 2L
Therefore the variation of shear force is parabolic
wL 6
at x = 0,
Vx =
at x = L,
Vx = 
wL 3 wL wx 2 x .x . 6 2L 3
and Bending Moment (Mx )
wL wx 3 .x 6 6L
The variation of BM is cubic at x = 0, Mx = 0 at x = L, Mx = 0 For maximum BM;
or
wL wx 2 6 2L
and Mmax
i.e.
d Mx
0
dx
0 or x
i.e. Vx
L 3
wL § L · w § L · u¨ u¨ ¸ ¸ 6 © 3 ¹ 6L © 3 ¹
Mmax
ª d Mx
« dx ¬
0
3
wL2 9 3
wL2 at x
9 3 Page 144 of 429
L 3
º Vx » ¼
S K Mondal’s
1 w x . x 2 L
wx 2 2L
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
(xiii) A Simply supported beam with a gradually varying load (GVL) zero at each end and w/unit length at mid span.
Consider equilibrium of the beam AB total load on the beam
Therefore R A
RB
§1 L · 2u¨ u u w¸ ©2 2 ¹
wL 4
The free body diagram of section A –XX as shown below, load at section XX (wx)
The resultant of that part of the distributed load which acts on this free body is applied at a point, distance x/3 from section XX.
Shear force (Vx):
wL 2
Page 145 of 429
2w .x L
1 2w .x. .x 2 L
wx 2 L
Chapter4
Bending Moment and Shear Force Diagram
In the region 0 < x < L/2
Vx
RA
wx 2 L
wL wx 2 4 L
Therefore the variation of shear force is parabolic.
wL 4
at x = 0,
Vx =
at x = L/4,
Vx = 0
In the region of L/2 < x < L The Diagram will be Mirror image of AC.
Bending moment (Mx): In the region 0 < x < L/2
wL § 1 2wx · .x ¨ .x. . x / 3
4 L ¸¹ ©2
Mx
wL wx 3 4 3L
The variation of BM is cubic at x = 0,
Mx = 0
at x = L/2, Mx =
wL2 12
In the region L/2 < x < L BM diagram will be mirror image of AC.
For maximum bending moment
d Mx
dx
or
0
wL wx 2 4 L
and Mmax
i.e.
i.e. Vx
ª d Mx
« dx ¬
0
0 or x
º Vx » ¼
L 2
wL2 12
Mmax
wL2 12
at x
Page 146 of 429
L 2
S K Mondal’s
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
(xiv) A Simply supported beam with a gradually varying load (GVL) zero at mid span and w/unit length at each end.
We now superimpose two beams as (1) Simply supported beam with a UDL through at its length
Vx 1 Mx 1
wL wx 2 wL wx 2 .x 2 2
And (2) a simply supported beam with a gradually varying load (GVL) zero at each end and w/unit length at mind span. In the range 0 < x < L/2
Vx 2 Mx 2
wL wx 2 4 L wL wx 3 .x 4 3L
Now superimposing we get
Shear force (Vx): In the region of 0< x < L/2 Page 147 of 429
Chapter4
Vx
Bending Moment and Shear Force Diagram Â§ wL wx 2 Â· wL Vx 1 Vx 2 Â§Â¨ wx Â·Â¸ Â¨ Â¸ L Â¹ Â© 2 Â¹ Â© 4
S K Mondalâ€™s
w 2 x  L/2
L Therefore the variation of shear force is parabolic at x = 0,
Vx = +
at x = L/2,
Vx = 0
wL 4
In the region L/2 < x < L The diagram will be mirror image of AC
Bending moment (Mx) = Mx 1  Mx 2 =
Â§ wL wx 2 Â· Â§ wL wx 3 Â· .x .x Â¨ Â¸ Â¨ Â¸ 2 Â¹ Â© 4 3L Â¹ Â© 2
wx 3 wx 2 wL .x 3L 2 4
The variation of BM is cubic
at x
0, Mx
0
at x
L / 2, Mx
wx 2 24
(xv) A simply supported beam with a gradually varying load (GVL) w1/unit length at one end and w2/unit length at other end.
Page 148 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
At first we will treat this problem by considering a UDL of identifying (w1)/unit length over the whole length and a varying load of zero at one end to (w2 w1)/unit length at the other end. Then superimpose the two loadings.
Consider a section XX at a distance x from left end A (i) Simply supported beam with UDL (w1) over whole length
Vx 1 Mx 1
w1L w1x 2 w1L 1 .x w1x 2 2 2
And (ii) simply supported beam with (GVL) zero at one end (w2 w1) at other end gives
Vx 2
w 2 w1 w 2 w1 x 2 6
M x 2 w 2 w 1 .
L .x 6
2L w 2 w1 x 3 6L
Now superimposing we get
Shear force Vx
Vx 1 + Vx 2
w1L w 2L x2 + w 1x w 2 w1
3 6 2L
? The SF variation is parabolic at x
0, Vx
at x L,
Vx
Bending moment Mx
w1L w 2L L 2w1 w 2
3 6 6 L w1 2w 2
6
Mx 1 Mx 2
w1L wL 1 Â§ w w Âˇ .x 1 .x w1x 2 Â¨ 2 1 Â¸ .x 3 3 6 2 ÂŠ 6L Âš
?The BM variation is cubic.
at x
0,
Mx
0
at x
L,
Mx
0
Page 149 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
(xvi) A Simply supported beam carrying a continuously distributed load. The intensity of the load at any point is, w x
Â§Sx Â· w sin Â¨ Â¸ . Where â€˜xâ€™ is the distance from each end of Â© L Â¹
the beam.
We will use Integration method as it is easier in this case. We know that
Therefore
d Vx
dx
d Vx
dx d Vx
load
and
d Mx
dx
Vx
Â§Sx Â· w sin Â¨ Â¸ Â© L Â¹ Â§Sx Â· w sin Â¨ Â¸ dx Â© L Â¹
Integrating both side we get
Â³ dV
x
Â§Sx Â· w Â³ sin Â¨ Â¸ dx Â© L Â¹
or
Vx
Â§SxÂ· w cos Â¨ Â¸ Â© L Â¹ A
S
L
> where, A
constant of Integration@
Page 150 of 429
Â§SxÂ· cos Â¨ Â¸A S Â© L Â¹
wL
Chapter4 Again we know that d Mx
dx
Vx
Bending Moment and Shear Force Diagram
or
d Mx
Vx dx
S K Mondalâ€™s
Â wL Â½ Â§Sx Â· cos Â¨ A Â¾ dx Â® Â¸ Â© L Â¹ Â¯S Â¿
Integrating both side we get
Â§Sx Â· sin Â¨ Â¸ S Â© L Â¹ Ax + B
wL Mx
wL2
S
S2
Â§Sx Â· sin Â¨ Â¸ Ax + B Â© L Â¹
L [Where B = constant of Integration] Now apply boundary conditions At x = 0,
Mx = 0
and
at x = L,
Mx = 0
This gives A = 0 and B = 0
Â§Sx Â· cos Â¨ Â¸ S Â© L Â¹ wL2 Â§Sx Â· sin Â¨ Â¸ 2 S Â© L Â¹
? Shear force Vx
And Mx
?
Mmax
wL2
S2
wL
and
Vmax
wL
S
at x
0
at x = L/2
(xvii) A Simply supported beam with a couple or moment at a distance â€˜aâ€™ from left end.
Considering equilibrium we get Page 151 of 429
Chapter4
¦M
A
Bending Moment and Shear Force Diagram 0 gives
RB ×L +M 0 or RB and
¦M
B
S K Mondal’s
M L
0 gives M L
R A ×L +M 0 or R A
Now consider any crosssection XX which is at a distance ‘x’ from left end A and another section YY at a distance ‘x’ from left end A as shown in figure.
In the region 0 < x < a Shear force (Vx) = RA =
M L
Bending moment (Mx) = RA.x =
M .x L
In the region a< x < L Shear force (Vx) = RA =
M L
Bending moment (Mx) = RA.x – M =
M .x  M L
Page 152 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
(xviii) A Simply supported beam with an eccentric load
When the beam is subjected to an eccentric load, the eccentric load is to be changed into a couple = Force u (distance travel by force) = P.a (in this case) and a force = P Therefore equivalent load diagram will be
Considering equilibrium
ÂŚM
A
0 gives
P.(L/2) + P.a + RB u L = 0 or RB =
P P.a 2 L
and RA + RB = P gives RA =
P P.a 2 L
Now consider any crosssection XX which is at a distance â€˜xâ€™ from left end A and another section YY at a distance â€˜xâ€™ from left end A as shown in figure.
Page 153 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
In the region 0 < x < L/2 Shear force (Vx) =
P P.a 2 L Â§ P Pa Âˇ Â¸.x ÂŠ2 L Âš
Bending moment (Mx) = RA . x = Â¨
In the region L/2 < x < L Shear force (Vx) =
P Pa P 2 L
=
P Pa 2 L
Bending moment (Vx) = RA . x â€“ P.( x  L/2 ) â€“ M =
PL Â§ P Pa Âˇ .x  Pa 2 Â¨ÂŠ 2 L Â¸Âš
4.6 Bending Moment diagram of Statically Indeterminate beam Beams for which reaction forces and internal forces cannot be found out from static equilibrium equations alone are called statically indeterminate beam. This type of beam requires deformation equation in addition to static equilibrium equations to solve for unknown forces.
Statically determinate  Equilibrium conditions sufficient to compute reactions. Statically indeterminate  Deflections (Compatibility conditions) along with equilibrium equations should be used to find out reactions. Page 154 of 429
Chapter4 Bending Moment and Shear Force Diagram S K Mondalâ€™s Type of Loading & B.M Diagram Reaction Bending Moment
R A= R B =
RA = RB =
P 2
wL 2
wL2 M =M = 12 A
B
RA
Pab 2 L2
RB
Pa 2 (3b a) L3
MB = 
Pa 2b L2
3wL 16
5wL 8
RB
+ 
B
MA = 
Rc =

A
Pb 2 (3a b) L3
R A= R B =
RA
PL M =M = 8
Page 155 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
4.7 Load and Bending Moment diagram from Shear Force diagram OR Load and Shear Force diagram from Bending Moment diagram If S.F. Diagram for a beam is given, then (i)
If S.F. diagram consists of rectangle then the load will be point load
(ii)
If S.F diagram consists of inclined line then the load will be UDL on that portion
(iii)
If S.F diagram consists of parabolic curve then the load will be GVL
(iv)
If S.F diagram consists of cubic curve then the load distribute is parabolic. After finding load diagram we can draw B.M diagram easily.
If B.M Diagram for a beam is given, then (i)
If B.M diagram consists of inclined line then the load will be free point load
(ii)
If B.M diagram consists of parabolic curve then the load will be U.D.L.
(iii)
If B.M diagram consists of cubic curve then the load will be G.V.L.
(iv)
If B.M diagram consists of fourth degree polynomial then the load distribution is parabolic.
Let us take an example: Following is the S.F diagram of a beam is given. Find its loading diagram.
Answer: From AE inclined straight line so load will be UDL and in AB = 2 m length load = 6 kN if UDL is w N/m then w.x = 6 or w u 2 = 6 or w = 3 kN/m after that S.F is constant so no force is there. At last a 6 kN for vertical force complete the diagram then the load diagram will be
As there is no support at left end it must be a cantilever beam. Page 156 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
4.8 Point of Contraflexure In a beam if the bending moment changes sign at a point, the point itself having zero bending moment, the beam changes curvature at this point of zero bending moment and this point is called the point of contra flexure. Consider a loaded beam as shown below along with the B.M diagrams and deflection diagram.
In this diagram we noticed that for the beam loaded as in this case, the bending moment diagram is partly positive and partly negative. In the deflected shape of the beam just below the bending moment diagram shows that left hand side of the beam is ‘sagging' while the right hand side of the beam is ‘hogging’. The point C on the beam where the curvature changes from sagging to hogging is a point of contraflexure.
x
There can be more than one point of contraflexure in a beam.
4.9 General expression x
d4y EI dx 2
Z
x
EI
d3y dx3
Vx
x
d2y EI 2 dx
Mx Page 157 of 429
Chapter4
Bending Moment and Shear Force Diagram x
dy = Č™ = slope dx
x
y= G = Deflection
x
Flexural rigidity = EI
Page 158 of 429
S K Mondalâ€™s
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20Years GATE Questions Shear Force (S.F.) and Bending Moment (B.M.) GATE1.
A concentrated force, F is applied (perpendicular to the plane of the figure) on the tip of the bent bar shown in Figure. The equivalent load at a section close to the fixed end is: (a) Force F (b) Force F and bending moment FL (c) Force F and twisting moment FL (d) Force F bending moment F L, and twisting moment FL [GATE1999]
GATE1. Ans. (c) GATE2.
The shear force in a beam subjected to pure positive bending is…… (positive/zero/negative) [GATE1995] GATE2. Ans. Zero
Cantilever GATE3.
Two identical cantilever beams are supported as shown, with their free ends in contact through a rigid roller. After the load P is applied, the free ends will have [GATE2005]
(a) (b) (c) (d)
Equal deflections but not equal slopes Equal slopes but not equal deflections Equal slopes as well as equal deflections Neither equal slopes nor equal deflections
GATE3. Ans. (a) As it is rigid roller, deflection must be same, because after deflection they also will be in contact. But slope unequal. GATE4.
A beam is made up of two identical bars AB and BC, by hinging them together at B. The end A is builtin (cantilevered) and the end C is simplysupported. With the load P acting as shown, the bending moment at A is: [GATE2005] Page 159 of 429
Chapter4
Bending Moment and Shear Force Diagram (a) Zero
(b)
PL 2
(c)
3PL 2
S K Mondalâ€™s (d) Indeterminate
GATE4. Ans. (b)
Cantilever with Uniformly Distributed Load GATE5.
The shapes of the bending moment diagram for a uniform cantilever beam carrying a uniformly distributed load over its length is: [GATE2001] (a) A straight line (b) A hyperbola (c) An ellipse (d) A parabola GATE5. Ans. (d)
Cantilever Carrying load Whose Intensity varies GATE6.
A cantilever beam carries the antisymmetric load shown, where Çšo is the peak intensity of the distributed load. Qualitatively, the correct bending moment diagram for this beam is:
[GATE2005]
GATE6. Ans. (d)
Page 160 of 429
Chapter4
Bending Moment and Shear Force Diagram
Mx
S K Mondalâ€™s
wx 2 wx 3 2 6L
Simply Supported Beam Carrying Concentrated Load GATE7.
A concentrated load of P acts on a simply supported beam of span L at a distance
L from the left support. The bending moment at the point of 3
application of the load is given by
PL (a) 3
2 PL (b) 3
[GATE2003]
PL (c ) 9
2 PL (d ) 9
GATE7. Ans. (d) Mc
GATE8.
Pab l
Â§ L Âˇ Â§ 2L Âˇ PuÂ¨ Â¸uÂ¨ Â¸ ÂŠ3Âš ÂŠ 3 Âš L
2PL 9
A simply supported beam carries a load 'P' through a bracket, as shown in Figure. The maximum bending moment in the beam is (a) PI/2 (b) PI/2 + aP/2 (c) PI/2 + aP (d) PI/2 â€“ aP [GATE2000]
GATE8. Ans. (c) Taking moment about Ra P u l Pa Rbl 0 2 P a P a or Rb P ?Ra P 2 l 2 l Maximum bending moment will be at centre â€˜Câ€™ l l Pl ?Mc Ra u P u a Rb u Pa or Mmax 2 2 2
Simply Supported Distributed Load
Beam
Carrying
a
Uniformly
Statement for Linked Answer and Questions Q9Q10: A mass less beam has a loading pattern as shown in the figure. The beam is of rectangular Page 161 of 429 crosssection with a width of 30 mm and height of 100 mm. [GATE2010]
Chapter4
Bending Moment and Shear Force Diagram
The maximum bending moment occurs at (a) Location B (c) 2500 mm to the right of A GATE9. Ans. (C)
S K Mondalâ€™s
GATE9.
(b) 2675 mm to the right of A (d) 3225 mm to the right of A
3000 N/m
R1 R1 R 2
R2 3000 u 2
R1 u 4 3000 u 2 u 1 R1
6000N 0
1500,
S.F. eqn . at any section x from end A. R1 3000 u x 2
x
^for
0
x ! 2m}
2.5 m.
GATE10. The maximum magnitude of bending stress (in MPa) is given by (a) 60.0 (b) 67.5 (c) 200.0 (d) 225.0 GATE10. Ans. (b) Binding stress will be maximum at the outer surface So taking 9 = 50 mm
ld 3 12
m u 50 3 ld 12 x2 m x 1.5 u 103 [2000 x] 2 6 ? m2500 3.375 u10 N mm and I
?V
& V
3.375 u 106 u 50 u 12 30 u 1003
67.5 MPa
Data for Q11Q12 are given below. Solve the problems and choose correct answers A steel beam of breadth 120 mm and height 750 mm is loaded as shown in the figure. Assume Esteel= 200 GPa.
[GATE2004] GATE11. The beam is subjected to a maximum bending moment of (a) 3375 kNm (b) 4750 kNm (c) 6750 kNm Page 162 of 429
(d) 8750 kNm
Chapter4
Bending Moment and Shear Force Diagram 2
GATE11. Ans. (a) Mmax
wl 8
120 u 15 kNm 8
3375kNm
GATE12. The value of maximum deflection of the beam is: (a) 93.75 mm (b) 83.75 mm (c) 73.75 mm 3
5 wl4 384 EI
0.12 u 0.75
(d) 63.75 mm
3
bh 4.22 u 103 m4 12 12 5 120 u 103 u 154 u m 93.75mm 384 200 u 109 u 4.22 u 103
GATE12. Ans. (a) Moment of inertia (I) =
G max
S K Mondal’s
2
Statement for Linked Answer and Questions Q13Q14: A simply supported beam of span length 6m and 75mm diameter carries a uniformly distributed load of 1.5 kN/m [GATE2006] GATE13. What is the maximum value of bending moment? (a) 9 kNm (b) 13.5 kNm (c) 81 kNm (d) 125 kNm 2 2 wl 1.5 u 6 6.75kNm But not in choice. Nearest choice (a) GATE13. Ans. (a) Mmax 8 8 GATE14. What is the maximum value of bending stress? (a) 162.98 MPa (b) 325.95 MPa (c) 625.95 Mpa 3 32M 32 u 6.75 u 10 GATE14. Ans. (a) V Pa 162.98MPa 2 S d3 S u 0.075
(d) 651.90 Mpa
Simply Supported Beam Carrying a Load whose Intensity varies Uniformly from Zero at each End to w per Unit Run at the MiD Span GATE15. A simply supported beam is subjected to a distributed loading as shown in the diagram given below: What is the maximum shear force in the beam? (a) WL/3 (b) WL/2 (c) WL/3 (d) WL/6 [IES2004] 1 WL uL u W Total load 2 2 § · ¸ WL Wx 2 WL 1 ¨ W x. ¨ u X ¸ Sx 4 2 ¨ L 4 L ¸ ¨ ¸ © 2 ¹ WL Smax at x 0 4
GATE15. Ans. (d)
GATE16. A simply supported beam of length 'l' is subjected to a symmetrical uniformly varying load with zero intensity at the ends and intensity w (load per unit length) at the mid span. What is the maximum bending moment? [IAS2004] Page 163 of 429
Chapter4
Bending Moment and Shear Force Diagram (a)
3wl 8
2
2
(b)
S K Mondalâ€™s
2
wl 12
(c)
wl 24
(d)
5wl 2 12
GATE16. Ans. (b)
Previous 20Years IES Questions Shear Force (S.F.) and Bending Moment (B.M.) IES1.
A lever is supported on two hinges at A and C. It carries a force of 3 kN as shown in the above figure. The bending moment at B will be (a) 3 kNm (b) 2 kNm (c) 1 kNm (d) Zero
[IES1998] IES1. Ans. (a) IES2.
A beam subjected to a load P is shown in the given figure. The bending moment at the support AA of the beam will be (a) PL (b) PL/2 (c) 2PL (d) zero
[IES1997] IES2. Ans. (b) Load P at end produces moment
PL in 2
anticlockwise direction. Load P at end produces moment of PL in clockwise direction. Net moment at AA is PL/2.
IES3.
The bending moment (M) is constant over a length segment (I) of a beam. The shearing force will also be constant over this length and is given by [IES1996] (a) M/l (b) M/2l (c) M/4l (d) None of the above IES3. Ans. (d) Dimensional analysis gives choice (d) IES4.
A rectangular section beam subjected to a bending moment M varying along its length is required to develop same maximum bending stress at any crosssection. If the depth of the section is constant, then its width will vary as [IES1995] (a) M
IES4. Ans. (a) IES5.
M I
(b)
M
(c) M2
(d) 1/M
3
const.
and
I
bh 12
Consider the following statements: [IES1995] If at a section distant from one of the ends of the beam, M represents the bending moment. V the shear force and w the intensity of loading, then 1. dM/dx = V 2. dV/dx = w 3. dw/dx = y (the deflection of the beam at the section) Select the correct answer using the codes given below: (a) 1 and 3 (b) 1 and 2 (c) 2 and 3 (d) 1, 2 and 3 Page 164 of 429 IES5. Ans. (b)
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
Cantilever IES6.
The given figure shows a beam BC simply supported at C and hinged at B (free end) of a cantilever AB. The beam and the cantilever carry forces of
100 kg and 200 kg respectively. The bending moment at B is: [IES1995] (a) Zero (b) 100 kgm (c) 150 kgm (d) 200 kgm IES6. Ans. (a) IES7.
Match ListI with ListII and select the correct answer using the codes given below the lists: [IES1993] ListI ListII (Condition of beam) (Bending moment diagram) A. Subjected to bending moment at the 1. Triangle end of a cantilever B. Cantilever carrying uniformly distributed 2. Cubic parabola load over the whole length C. Cantilever carrying linearly varying load 3. Parabola from zero at the fixed end to maximum at the support D. A beam having load at the centre and 4. Rectangle supported at the ends Codes: A B C D A B C D (a) 4 1 2 3 (b) 4 3 2 1 (c) 3 4 2 1 (d) 3 4 1 2 IES7. Ans. (b) IES8.
If the shear force acting at every section of a beam is of the same magnitude and of the same direction then it represents a [IES1996] (a) Simply supported beam with a concentrated load at the centre. (b) Overhung beam having equal overhang at both supports and carrying equal concentrated loads acting in the same direction at the free ends. (c) Cantilever subjected to concentrated load at the free end. (d) Simply supported beam having concentrated loads of equal magnitude and in the same direction acting at equal distances from the supports. IES8. Ans. (c)
Cantilever with Uniformly Distributed Load A uniformly distributed load Z (in kN/m) is acting over the entire length of a 3 m long cantilever beam. If the shear force at the midpoint of cantilever is 6 kN, what is the value of Z ? [IES2009] (a) 2 (b) 3 (c) 4 (d) 5 IES9. Ans. (c) IES9.
Shear force at mid point of cantilever
Zl 2
Zu 3 2
Z
6
6 6u2 3
4 kN / m Page 165 of 429
Chapter4 IES10.
Bending Moment and Shear Force Diagram
Code: (a) (c) IES10. Ans. (b) IES11.
S K Mondalâ€™s
Match ListI with ListII and select the correct answer using the code given below the Lists: [IES2009]
A 1 1
B 5 3
C 2 4
D 4 5
(b) (d)
A 4 4
B 5 2
C 2 5
D 3 3
The shearing force diagram for a beam is shown in the above figure. The bending moment diagram is represented by which one of the following?
[IES2008]
IES11. Ans. (b) Uniformly distributed load on cantilever beam.
Page 166 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
IES12.
A cantilever beam having 5 m length is so loaded that it develops a shearing force of 20T and a bending moment of 20 Tm at a section 2m from the free end. Maximum shearing force and maximum bending moment developed in the beam under this load are respectively 50 T and 125 Tm. The load on the beam [IES1995] is: (a) 25 T concentrated load at free end (b) 20T concentrated load at free end (c) 5T concentrated load at free end and 2 T/m load over entire length (d) 10 T/m udl over entire length IES12. Ans. (d)
Cantilever Carrying Uniformly Distributed Load for a Part of its Length IES13.
A vertical hanging bar of length L and weighing w N/ unit length carries a load W at the bottom. The tensile force in the bar at a distance Y from the support will be given by [IES1992]
a W wL
b W w( L y )
c W w y / L
d W
W (L y) w
IES13. Ans. (b)
Cantilever Carrying load Whose Intensity varies IES14.
A cantilever beam of 2m length supports a triangularly distributed load over its entire length, the maximum of which is at the free end. The total load is 37.5 kN.What is the bending moment at the fixed end? [IES 2007] (a) 50 u 106 N mm (b) 12.5 u 106 N mm (c) 100 u 106 N mm (d) 25 u 106 N mm IES14. Ans. (a)
Page 167 of 429
Chapter4
Bending Moment and Shear Force Diagram
M = 37.5 u
S K Mondalâ€™s
4 KNm = 50 u 106 Nmm 3
Simply Supported Beam Carrying Concentrated Load IES15.
Assertion (A): If the bending moment along the length of a beam is constant, then the beam cross section will not experience any shear stress. [IES1998] Reason (R): The shear force acting on the beam will be zero everywhere along the length. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES15. Ans. (a) IES16. Assertion (A): If the bending moment diagram is a rectangle, it indicates that the beam is loaded by a uniformly distributed moment all along the length. Reason (R): The BMD is a representation of internal forces in the beam and not the moment applied on the beam. [IES2002] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES16. Ans. (d) IES17.
The maximum bending moment in a simply supported beam of length L loaded by a concentrated load W at the midpoint is given by [IES1996] (a) WL
(b)
WL 2
(c)
WL 4
(d)
WL 8
IES17. Ans. (c) IES18.
A simply supported beam is loaded as shown in the above figure. The maximum shear force in the beam will be (a) Zero (b) W (c) 2W (d) 4W [IES1998]
IES18. Ans. (c) IES19.
If a beam is subjected to a constant bending moment along its length, then the shear force will [IES1997] (a) Also have a constant value everywhere along its length (b) Be zero at all sections along the beam (c) Be maximum at the centre and zero at the ends (d) zero at the centre and maximum at the ends IES19. Ans. (b) Page 168 of 429
Chapter4 IES20.
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
A loaded beam is shown in the figure. The bending moment diagram of the beam is best represented as:
[IES2000]
IES20. Ans. (a) IES21.
A simply supported beam has equal overhanging lengths and carries equal concentrated loads P at ends. Bending moment over the length between the supports [IES2003] (a) Is zero (b) Is a nonzero constant (c) Varies uniformly from one support to the other (d) Is maximum at midspan IES21. Ans. (b)
IES22.
The bending moment diagram for the case shown below will be q as shown in
(a)
(b)
(c)
(d)
[IES1992] IES22. Ans. (a)
Page 169 of 429
Chapter4 IES23.
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
Which one of the following portions of the loaded beam shown in the given figure is subjected to pure bending? (a) AB (b)DE (c) AE (d) BD [IES1999]
IES23. Ans. (d) Pure bending takes place in the section between two weights W IES24.
Constant bending moment over span "l" will occur in
[IES1995]
IES24. Ans. (d) IES25. For the beam shown in the above figure, the elastic curve between the supports B and C will be: (a) Circular (b) Parabolic (c) Elliptic (d) A straight line
[IES1998] IES25. Ans. (b) IES26.
A beam is simply supported at its ends and is loaded by a couple at its midspan as shown in figure A. Shear force diagram for the beam is given by the figure. [IES1994]
(a) B IES26. Ans. (d)
IES27.
(b) C
(c) D
(d) E
A beam AB is hingedsupported at its ends and is loaded by couple P.c. as shown in the given figure. The magnitude or shearing force at a section x of the [IES1993] beam is:
(a) 0
(b) P
(c) P/2L
Page 170 of 429
(d) P.c./2L
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
IES27. Ans. (d) If F be the shearing force at section x (at point A), then taking moments about B, F x 2L = Pc
or F
Pc 2L
Thus shearing force in zone x
Simply Supported Distributed Load IES28.
Beam
Pc 2L
Carrying
a
Uniformly
A freely supported beam at its ends carries a central concentrated load, and maximum bending moment is M. If the same load be uniformly distributed over the beam length, [IES2009] then what is the maximum bending moment? (a) M
(b)
M 2
(c)
IES28. Ans. (b)
B.MMax
WL 4
M
Where the Load is U.D.L. Maximum Bending Moment 2 § W ·§ L · ¨ ¨ L ¸ 8¸ © ¹© ¹ WL 1 § WL · 8 2 ¨© 4 ¸¹
M 2
Page 171 of 429
M 3
(d) 2M
Chapter4
Bending Moment and Shear Force Diagram
S K Mondal’s
Simply Supported Beam Carrying a Load whose Intensity varies Uniformly from Zero at each End to w per Unit Run at the MiD Span IES29.
A simply supported beam is subjected to a distributed loading as shown in the diagram given below: What is the maximum shear force in the beam? (a) WL/3 (b) WL/2 (c) WL/3 (d) WL/6 [IES2004] 1 WL uL u W Total load 2 2 § · ¸ WL Wx 2 WL 1 ¨ W x. ¨ u X ¸ Sx 4 2 ¨ L 4 L ¸¸ ¨ © 2 ¹ WL Smax at x 0 4
IES29. Ans. (d)
Simply Supported Beam carrying a Load whose Intensity varies IES30.
A beam having uniform crosssection carries a uniformly distributed load of intensity q per unit length over its entire span, and its midspan deflection is ǅ.
The value of midspan deflection of the same beam when the same load is distributed with intensity varying from 2q unit length at one end to zero at the other end is: [IES1995] (a) 1/3 ǅ (b) 1/2 ǅ (c) 2/3 ǅ (d) ǅ IES30. Ans. (d)
Simply Supported Beam with Equal Overhangs and carrying a Uniformly Distributed Load IES31.
A beam, builtin at both ends, carries a uniformly distributed load over its entire span as shown in figureI. Which one of the diagrams given below, represents bending moment distribution along the length of the beam? [IES1996]
Page 172 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
IES31. Ans. (d)
The Points of Contraflexure IES32.
The pointÂˇ of contraflexure is a point where: [IES2005] (a) Shear force changes sign (b) Bending moment changes sign (c) Shear force is maximum (d) Bending moment is maximum IES32. Ans. (b) IES33.
Match List I with List II and select the correct answer using the codes given below the Lists: [IES2000] ListI ListII A. Bending moment is constant 1. Point of contraflexure B. Bending moment is maximum or minimum 2. Shear force changes sign C. Bending moment is zero 3. Slope of shear force diagram is zero over the portion of the beam D. Loading is constant 4. Shear force is zero over the portion of the beam Code: A B C D A B C D (a) 4 1 2 3 (b) 3 2 1 4 (c) 4 2 1 3 (d) 3 1 2 4 IES33. Ans. (b)
Loading and B.M. diagram from S.F. Diagram IES34.
The bending moment diagram shown in Fig. I correspond to the shear force diagram in [IES1999]
IES34. Ans. (b) If shear force is zero, B.M. will also be zero. If shear force varies linearly with length, B.M. diagram will be curved line. IES35.
Bending moment distribution in a built be am is shown in the given Page 173 of 429
Chapter4
Bending Moment and Shear Force Diagram
The shear force distribution in the beam is represented by
S K Mondalâ€™s
[IES2001]
IES35. Ans. (a) IES36.
The given figure shows the shear force diagram for the beam ABCD. Bending moment in the portion BC of the beam [IES1996] (b) Is zero (d) Varies parabolically from B to C
(a) Is a nonzero constant (c) Varies linearly from B to C IES36. Ans. (a)
IES37.
Figure shown above represents the BM diagram for a simply supported beam. The beam is subjected to which one of the following? (a) A concentrated load at its midlength (b) A uniformly distributed load over its length (c) A couple at its midlength (d) Couple at 1/4 of the span from each end [IES2006]
IES37. Ans. (c) IES38.
If the bending moment diagram for a simply supported beam is of the form given below. Then the load acting on the beam is: (a) A concentrated force at C (b) A uniformly distributed load over the whole length of the beam (c) Equal and opposite moments applied at A and B (d) A moment applied at C [IES1994] IES38. Ans. (d) A vertical line in centre of B.M. diagram is possible when a moment is applied there. IES39.
The figure given below shows a bending moment diagram for the beam CABD: Page 174 of 429
Chapter4
Bending Moment and Shear Force Diagram
Load diagram for the above beam will be:
S K Mondalâ€™s
[IES1993]
IES39. Ans. (a) Load diagram at (a) is correct because B.M. diagram between A and B is parabola which is possible with uniformly distributed load in this region. IES40.
The shear force diagram shown in the following figure is that of a [IES1994] (a) Freely supported beam with symmetrical point load about midspan. (b) Freely supported beam with symmetrical uniformly distributed load about midspan (c) Simply supported beam with positive and negative point loads symmetrical about the midspan (d) Simply supported beam with symmetrical varying load about midspan
IES40. Ans. (b) The shear force diagram is possible on simply supported beam with symmetrical varying load about mid span.
Previous 20Years IAS Questions Shear Force (S.F.) and Bending Moment (B.M.) IAS1.
Assertion (A): A beam subjected only to end moments will be free from shearing force. [IAS2004] Reason (R): The bending moment variation along the beam length is zero. Page 175 of 429 (a) Both A and R are individually true and R is the correct explanation of A
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
(b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS1. Ans. (a) IAS2. Assertion (A): The change in bending moment between two crosssections of a beam is equal to the area of the shearing force diagram between the two sections. [IAS1998] Reason (R): The change in the shearing force between two crosssections of beam due to distributed loading is equal to the area of the load intensity diagram between the two sections. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS2. Ans. (b)
IAS3.
The ratio of the area under the bending moment diagram to the flexural rigidity between any two points along a beam gives the change in [IAS1998] (a) Deflection (b) Slope (c) Shear force (d) Bending moment IAS3. Ans. (b)
Cantilever IAS4.
A beam AB of length 2 L having a concentrated load P at its midspan is hinge supported at its two ends A and B on two identical cantilevers as shown in the given figure. The correct value of bending moment at A is (a) Zero (b) PLl2 [IAS1995] (c) PL (d) 2 PL IAS4. Ans. (a) Because of hinge support between beam AB and cantilevers, the bending moment can't be transmitted to cantilever. Thus bending moment at points A and B is zero. IAS5.
A load perpendicular to the plane of the handle is applied at the free end as shown in the given figure. The values of Shear Forces (S.F.), Bending Moment (B.M.) and torque at the fixed end of the handle have been determined respectively as 400 N, 340 Nm and 100 by a student. Among these values, those of [IAS1999] (a) S.F., B.M. and torque are correct (b) S.F. and B.M. are correct (c) B.M. and torque are correct (d) S.F. and torque are correct
IAS5. Ans. (d) S.F 400N and BM
Torque
400 u 0.25
400 u 0.4 0.2
240Nm
100Nm
Cantilever with Uniformly Distributed Load IAS6.
If the SF diagram for a beam is a triangle with length of the beam as its base, the beam is: [IAS2007] (a) A cantilever with a concentrated load at its free end (b) A cantilever with udl over its whole span (c) Simply supported with a concentrated load Page 176 of 429 at its midpoint (d) Simply supported with a udl over its whole span
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
IAS6. Ans. (b)
IAS7.
A cantilever carrying a uniformly distributed load is shown in Fig. I. Select the correct R.M. diagram of the cantilever. [IAS1999]
IAS7. Ans. (c) Mx
IAS8.
wx u
x 2
wx 2 2
A structural member ABCD is loaded as shown in the given figure. The shearing force at any section on the length BC of the member is: (a) Zero (b) P (c) Pa/k (d) Pk/a [IAS1996]
IAS8. Ans. (a)
Cantilever Carrying load Whose Intensity varies IAS9.
The beam is loaded as shown in Fig. I. Select the correct B.M. diagram [IAS1999]
Page 177 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
IAS9. Ans. (d)
Simply Supported Beam Carrying Concentrated Load IAS10.
Assertion (A): In a simply supported beam carrying a concentrated load at midspan, both the shear force and bending moment diagrams are triangular in nature without any change in sign. [IAS1999] Reason (R): When the shear force at any section of a beam is either zero or changes sign, the bending moment at that section is maximum. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS10. Ans. (d) A is false.
IAS11.
For the shear force to be uniform throughout the span of a simply supported beam, it should carry which one of the following loadings? [IAS2007] (a) A concentrated load at midspan (b) Udl over the entire span (c) A couple anywhere within its span (d) Two concentrated loads equal in magnitude and placed at equal distance from each support
Page 178 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
IAS11. Ans. (d) It is a case of pure bending.
IAS12.
Which one of the following figures represents the correct shear force diagram for the loaded beam shown in the given figure I? [IAS1998; IAS1995]
IAS12. Ans. (a)
Simply Supported Distributed Load IAS13.
Beam
Carrying
a
Uniformly
For a simply supported beam of length fl' subjected to downward load of uniform intensity w, match ListI with ListII and select the correct answer using the codes given below the Lists: [IAS1997] ListI ListII A.
Slope of shear force diagram
5wL 4 1. 384 E I
B.
Maximum shear force
2. w
C.
Maximum deflection
3.
D.
Magnitude of maximum bending moment
4.
Codes: (a) (c)
A 1 3
B 2 2
C 3 1
D 4 (b) Page 429 4 179 of (d)
A 3 2
wL 4 8 wL 2 B 1 4
C 2 1
D 4 3
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
IAS13. Ans. (d)
Simply Supported Beam Carrying a Load whose Intensity varies Uniformly from Zero at each End to w per Unit Run at the MiD Span IAS14.
A simply supported beam of length 'l' is subjected to a symmetrical uniformly varying load with zero intensity at the ends and intensity w (load per unit length) at the mid span. What is the maximum bending moment? [IAS2004]
3wl 2 (a) 8
wl 2 (b) 12
wl 2 (c) 24
5wl 2 (d) 12
IAS14. Ans. (b)
Simply Supported Beam carrying a Load whose Intensity varies IAS15.
A simply supported beam of span l is subjected to a uniformly varying load having zero intensity at the left support and w N/m at the right support. The reaction at the right support is: [IAS2003] (a)
wl 2
(b)
wl 5
(c)
wl 4
(d)
wl 3
IAS15. Ans. (d)
Simply Supported Beam with Equal Overhangs and carrying a Uniformly Distributed Load IAS16.
Consider the following statements for a simply supported beam subjected to a couple at its midspan: [IAS2004] 1. Bending moment is zero at the ends and maximum at the centre 2. Bending moment is constant over the entire length of the beam 3. Shear force is constant over the entire length of the beam 4. Shear force is zero over the entire length of the beam Which of the statements given above are correct? (a) 1, 3 and 4 (b) 2, 3 and 4 (c) 1 and 3 (d) 2 and 4 IAS16. Ans. (c)
Page 180 of 429
Chapter4
IAS17.
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
Match ListI (Beams) with ListII (Shear force diagrams) and select the correct answer using the codes given below the Lists: [IAS2001]
Codes: (a) (c) IAS17. Ans. (d)
A 4 1
B 2 4
C 5 3
D 3 5
(b) (d)
A 1 4
B 4 2
C 5 3
D 3 5
The Points of Contraflexure IAS18.
A point, along the length of a beam subjected to loads, where bending moment changes its sign, is known as the point of [IAS1996] (a) Inflexion (b) Maximum stress (c) Zero shear force (d) Contra flexure IAS18. Ans. (d) IAS19.
Assertion (A): In a loaded beam, if the shear force diagram is a straight line parallel to the beam axis, then the bending moment is a straight line inclined to the beam axis. [IAS 1994] Reason (R): When shear force at any section of a beam is zero or changes sign, the bending moment at that section is maximum. (a) Both A and R are individually true and R is the correct explanation of A Page 181 of 429 (b) Both A and R are individually true but R is NOT the correct explanation of A
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
(c) A is true but R is false (d) A is false but R is true IAS19. Ans. (b)
Loading and B.M. diagram from S.F. Diagram IAS20.
The shear force diagram of a loaded beam is shown in the following figure: The maximum Bending Moment of the beam is: (a) 16 kNm (b) 11 kNm (c) 28 kNm
(d) 8 kNm
[IAS1997] IAS20. Ans. (a)
IAS21.
The bending moment for a loaded beam is shown below:
[IAS2003]
The loading on the beam is represented by which one of the followings diagrams? (b) (a)
(c)
(d)
IAS21. Ans. (d) Page 182 of 429
Chapter4 IAS22.
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
Which one of the given bending moment diagrams correctly represents that of the loaded beam shown in figure? [IAS1997]
IAS22. Ans. (c) Bending moment does not depends on moment of inertia. The shear force diagram is shown
IAS23.
above
for
a
corresponding
loaded
beam.
bending
The
moment
diagram is represented by
[IAS2003]
IAS23. Ans. (a) IAS24.
The bending moment diagram for a simply supported beam is a rectangle over a larger portion of the span except near the supports. What type of load does the beam carry? [IAS2007] (a) A uniformly distributed symmetrical load over a larger portion of the span except near the supports (b) A concentrated load at midspan (c) Two identical concentrated loads equidistant from the supports and close to midpoint of the beam Page 183 of 429
Chapter4
Bending Moment and Shear Force Diagram (d)
S K Mondalâ€™s
Two identical concentrated loads equidistant from the midspan and close to supports
IAS24. Ans. (d)
Page 184 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
Previous Conventional Questions with Answers Conventional Question IES2005 Question:
A simply supported beam of length 10 m carries a uniformly varying load whose intensity varies from a maximum value of 5 kN/m at both ends to zero at the centre of the beam. It is desired to replace the beam with another simply supported beam which will be subjected to the same maximum 'bending momentâ€™ and â€˜shear force' as in the case of the previous one. Determine the length and rate of loading for the second beam if it is subjected to a uniformly distributed load over its whole length. Draw the variation of 'SF' and 'BM' in both the cases.
Answer:
X
5KN/m
5KN/m
B X
10m
RA
Total load on beam =5Ă—
RB
10 25 kN 2
25 12.5 kN 2 Take a section XX from B at a distance x. For 0 b x b 5 m we get rate of loading = RA RB
X a bx [as lineary varying] at x=0, X =5 kN / m and at x = 5, X 0 These two bounday condition gives a = 5 and b = 1 = X 5 x
We know that shear force(V),
dV X dx
or V =Â¨ Xdx = Â¨ (5 x )dx 5 x
at x = 0, F =12.5 kN (RB ) so c1 12.5 x2
12.5 2 It is clear that maximum S.F = 12.5 kN = V = 5x +
Page 185 of 429
x2
c1 2
Chapter4
Bending Moment and Shear Force Diagram
For a beam
S K Mondalâ€™s
dM V dx
x2 5x 2 x 3
12.5)dx =
12.5 x C2 2 2 6 at x = 0, M = 0 gives C2 0 or , M =Â¨ Vdx Â¨ (5 x
M = 12.5x  2.5x 2 x 3 / 6
for Maximum bending moment at
dM 0 dx
x2
12.5 0 2 or , x 2 10 x 25 0 or5x+
or , x 5 means at centre. So, Mmax 12.5 q 2.5 2.5 q 52 53 / 6 20.83 kNm X
A
XKNm B
L
RA
RB X
Now we consider a simply supported beam carrying uniform distributed load over whole length (X KN/m). Here R A RB
WL 2
S.F .at section XX WA Xx 2 12.5 kN
Vx
Vmax
B.M at section XX WA Wx 2 x 2 2 2 2 dM x WL X Â?Âž L ÂŹÂ WL qÂž ÂÂ 20.83 (ii ) dx 2 2 ÂžÂ&#x; 2 ÂŽ 8 Solving (i ) & (ii ) we get L=6.666m and X =3.75kN/m
Mx
Page 186 of 429
Chapter4
Bending Moment and Shear Force Diagram X
5KN/m
S K Mondalâ€™s
5KN/m
3.75kN/m
B 12.5KN/m
12.5KN/m
X
10m
RA
RB
12.5kN
6.666m
12.5kN
S.F.D
S.F.D 20.83KNm
Cubic parabola
B.M.D
Parabola
20.83kNm
Conventional Question IES1996 Question:
Answer:
A Uniform beam of length L is carrying a uniformly distributed load w per unit length and is simply supported at its ends. What would be the maximum bending moment and where does it occur? By symmetry each support reaction is equal i.e. RA=RB=
WA 2
B.M at the section xx is Mx=+
WA Wx 2 x 2 2
For the B.M to be maximum we have to
dM x 0 that gives. dx
WA Xx 0 + 2 or x= A i.e. at mid point. 2
Bending Moment Diagram
2
XA A X Â AÂŻ wA 2 And Mmax= q qÂĄ Â°
2 2 ÂĄÂ˘ 2 Â°Âą 2 8 Conventional Question AMIE1996 Question:
Calculate the reactions at A and D for the beam shown in figure. Draw the bending moment and shear force diagrams showing all important values.
Page 187 of 429
Chapter4
Answer:
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
Equivalent figure below shows an overhanging beam ABCDF supported by a roller support at A and a hinged support at D. In the figure, a load of 4 kN is applied through a bracket 0.5 m away from the point C. Now apply equal and opposite load of 4 kN at C. This will be equivalent to a anticlockwise couple of the value of (4 x 0.5) = 2 kNm acting at C together with a vertical downward load of 4 kN at C. Show U.D.L. (1 kN/m) over the port AB, a point load of 2 kN vertically downward at F, and a horizontal load of 2 3 kN as shown.
For reaction and A and D. Let ue assume RA= reaction at roller A. RDV vertically component of the reaction at the hinged support D, and Page 188 of 429
RDH horizontal component of the reaction at the hinged support D.
Chapter4 Obviously
Bending Moment and Shear Force Diagram RDH= 2 3 kN ( o )
S K Mondalâ€™s
In order to determine RA, takings moments about D, we get
Â§2 Â· R A u 6 2 u 1 1u 2 u Â¨ 2 2 Â¸ 2 4 u 2 Â©2 Â¹ or R A 3kN Also R A RDV
1u 2 4 2
or
5kN vetrically upward
RDV
8
R R
2
? Re action at D, RD
DV
Inclination with horizontal T
2
DH
tan1
52 2 3
5 2 3
2
6.08kN
55.30
S.F.Calculation : VF
2kN
VD
2 5
3kN
VC
34
1kN
VB
1kN
VA
1 1u 2
3kN
B.M.Calculation : MF
0
MD
2 u 1 2kNm
MC
ÂªÂ¬ 2 1 2 5 u 2ÂºÂ¼ 2
6kNm
The bending moment increases from 4kNm in i,e., 2 1 2 5 u 2
to 6kNm as shown MB MP
MA
2 1 2 2 5 2 4 u 2 2
4kNm
2Â· 1 Â§ 2 Â¨ 1 2 2 Â¸ 5 2 2 1 4 2 1 2 1u 1u 2Â¹ 2 Â© 2.5kNm 0
Conventional Question GATE1997 Question:
Construct the bending moment and shearing force diagrams for the beam shown in the figure.
Answer: Page 189 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
Calculation: First find out reaction at B and E. Taking moments, about B, we get
RE u 4.5 20 u 0.5 u or
RE
RB RE RB
50 u 3 40 u 5
55kN
Also, or
0.5 100 2
20 u 0.5 50 40
> RE
45kN
S.F. Calculation :
B.M.Calculation :
55kN@
VF
40kN
VE
40 55 15kN
VD
15 50
35kN
VB 35 45 10kN MG 0 MF
0
ME
40 u 0.5
MD
40 u 2 55 u 1.5
20kNm
2.5kNm
MC 40 u 4 55 u 3.5 50 u 2 67.5kNm The bending moment increases from 62.5kNm to 100. 0.5 MB 20 u 0.5 u 2.5kNm 2 Conventional Question GATE1996 Question:
Two bars AB and BC are connected by a frictionless hinge at B. The assembly is supported and loaded as shown in figure below. Draw the shear force and bending moment diagrams for the combined beam AC. clearly labelling the Page 190 of 429
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
important values. Also indicate your sign convention.
Answer:
There shall be a vertical reaction at hinge B and we can split the problem in two parts. Then the FBD of each part is shown below
Calculation: Referring the FBD, we get,
Fy
0, and R1 R2
200kN
From
ÂŚM
or
R2
?
R1
500 125kN 4 200 125 75kN
Again, R3
R1
75kN
and
B
0,100 u 2 100 u 3 R2 u 4
M 75 u 1.5 112.5kNm.
Conventional Question IES1998
Page 191 of 429
0
Chapter4
Bending Moment and Shear Force Diagram
S K Mondalâ€™s
Question:
A tube 40 mm outside diameter; 5 mm thick and 1.5 m long simply supported at 125 mm from each end carries a concentrated load of 1 kN at each extreme end. (i) Neglecting the weight of the tube, sketch the shearing force and bending moment diagrams; (ii) Calculate the radius of curvature and deflection at midspan. Take the modulus of elasticity of the material as 208 GN/m2
Answer:
(i) Given, d0
W
1kN; E
40 mm 0.04 m; di
208GN / m
2
d0 2t 2
40 2 u 5
30 mm
2
208 u 10 N / m ; l 1.5; a 125mm
Calculation: (ii) Radius of coordinate R As per bending equation:
M I
V
E y R EI or R i
M Here,M W u a 1u 103 u 0.125 125Nm I
S
d 64
4 0
d14
S ÂŞ 4 4 0.04 0.03 Âş 8.59 u 108 m4
Âź 64 ÂŹ Substituting the values in equation i , we get 208 u 108 u 8.59 u 108 125 Deflection at mid span : R
EI
d2 y dx 2
Mx
142.9 m
Wx W x Page Wx Wx Wa a 192 of 429
Wa
0.03 m;
0.125 m
Chapter4
Bending Moment and Shear Force Diagram Integrating, we get EI
dy dx
S K Mondalâ€™s
Wax C1
1 dy , 0 2 dx 1 Wal 0 Wa C1 or C1 ? 2 2 dy Wal EI ? Wax dx 2 Integrating again, we get When,
x
When
x 2 Wal x C2 2 2 a, y 0
Wa
EIy x
Wa3 Wa2l C2 2 2 Wa3 Wa2l C2 2 2 2 Wax Walx Âª Wa3 Wa2l Âº EIy Â« Â» 2 2 2 Â¼ Â¬ 2
?
0
or
? or
y
At mid span,i,e., x y
Wa Âª x 2 lx a2 al Âº Â» Â« EI Â¬ 2 2 2 2 Â¼ l/2 2 l u l / 2 a2 al Âº Wa Âª l / 2
Â« Â» EI Â« 2 2 2 2Â» Â¬ Â¼ 2 2 Wa Âª l a al Âº Â» Â« EI Â¬ 8 2 2 Â¼ Âª 1.52 0.1252 0.125 u 1.5 Âº 1u 1000 u 0.125 Â« Â» 208 u 109 u 8.59 u 10 8 Â¬ 8 2 2 Â¼
0.001366m 1.366mm It will be in upward direction
Conventional Question IES2001 Question:
What is meant by point of contraflexure or point of inflexion in a beam? Show the same for the beam given below:
17.5kN/m A
4M Answer:
20kN C
B 4M
D 2m
In a beam if the bending moment changes sign at a point, the point itself having zero bending moment, the beam changes curvature at this point of zero bending moment and this point is called the point of contra flexure.
Page 193 of 429
Chapter4
Bending Moment and Shear Force Diagram
17.5kN/m A
20kN C
4M
S K Mondalâ€™s
B 4M
D 2M
BMD From the bending moment diagram we have seen that it is between A & C. [If marks are more we should calculate exact point.]
Page 194 of 429
5.
Deflection of Beam
Theory at a Glance (for IES, GATE, PSU) 5.1 Introduction
x
We know that the axis of a beam deflects from its initial position under action of applied forces.
x
In this chapter we will learn how to determine the elastic deflections of a beam.
Selection of coordinate axes We will not introduce any other coordinate system. We use general coordinate axis as shown in the figure. This system will be followed in deflection of beam and in shear force and bending moment diagram. Here downward direction will be negative i.e. negative Yaxis. Therefore downward deflection of the beam will be treated as negative.
We use above Coordinate system
To determine the value of deflection of beam subjected to a given loading where we will use the formula, EI
d 2y dx 2
Mx .
Some books fix a coordinate axis as shown in the following figure. Here downward direction will be positive i.e. positive Yaxis. Therefore downward deflection of the beam will be treated as positive. As beam is generally deflected in downward directions and
this
coordinate
system
treats
downward
deflection is positive deflection.
Some books use above coordinate system
To determine the value of deflection of beam subjected to a given loading where we will use the formula, EI
d 2y dx 2
M x .
Why to calculate the deflections?
x
To prevent cracking of attached brittle materials
x
To make sure the structure not deflect severely and to â€œappearâ€? safe for its occupants
x
To help analyzing statically indeterminate structures Page 195 of 429
Chapter5
x
Deflection of Beam
S K Mondal’s
Information on deformation characteristics of members is essential in the study of vibrations of machines
Several methods to compute deflections in beam x
Double integration method (without the use of singularity functions)
x
Macaulay’s Method (with the use of singularity functions)
x
Moment area method
x
Method of superposition
x
Conjugate beam method
x
Castigliano’s theorem
x
Work/Energy methods
Each of these methods has particular advantages or disadvantages.
Methods to find deflection
Double integration
Geometrical
Moment area method
Energy Method
Conjugate beam method
Castiglian’s theorem
Virtual Work
Assumptions in Simple Bending Theory x
Beams are initially straight
x
The material is homogenous and isotropic i.e. it has a uniform composition and its mechanical properties are the same in all directions
x
The stressstrain relationship is linear and elastic
x
Young’s Modulus is the same in tension as in compression
x
Sections are symmetrical about the plane of bending
x
Sections which are plane before bending remain plane after bending
NonUniform Bending x
In the case of nonuniform bending of a beam, where bending moment varies from section to section, there will be shear force at each cross section which will induce shearing stresses
x
Also these shearing stresses cause warping (or outof plane distortion) of the cross section so that plane cross sections do not remain plane even after bending Page 196 of 429
Chapter5
Deflection of Beam
5.2 Elastic line or Elastic curve We have to remember that the differential equation of the elastic line is
2
d y EI 2 =M x dx Proof: Consider the following simply supported beam with UDL over its length.
From elementary calculus we know that curvature of a line (at point Q in figure)
1 R
d2 y dx 2
where R radius of curvature
3/2
ÂÂ° Â§ dy Âˇ2 Â˝Â° ÂŽ1 Â¨ Â¸ Âž ÂŻÂ° ÂŠ dx Âš ÂżÂ°
For small deflection, or
dy 0 dx
1 d2 y  R dx 2
Page 197 of 429
S K Mondalâ€™s
Chapter5 Deflection of Beam Bending stress of the beam (at point Q)
S K Mondal’s
Mx .y
Vx
EI From strain relation we get
H 1 x and H x R y 1 Mx ? R EI d2 y Mx Therefore dx 2 EI d2 y or EI Mx dx 2
Vx E
5.3 General expression From the equation EI
d2y dx 2
M x we may easily find out the following relations.
x
EI
d4y dx 4
Z Shear force density (Load)
x
EI
d3y dx3
Vx
x
EI
d2y dx 2
M x Bending moment
x
dy = ș = slope dx
x x
y = G = Deflection, Displacement Flexural rigidity = EI
Shear force
5.4 Double integration method (without the use of singularity functions)
³
Z dx
x
Vx =
x
Mx = Vx dx
x
EI
d2y dx 2
x
T
Slope
x
G
Deflection
³
Mx
1 M x dx EI ³
³ T dx
4step procedure to solve deflection of beam problems by double integration method Page 198 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s
Step 1: Write down boundary conditions (Slope boundary conditions and displacement boundary conditions), analyze the problem to be solved
Step 2: Write governing equations for, EI
d2y dx 2
Mx
Step 3: Solve governing equations by integration, results in expression with unknown integration constants
Step 4: Apply boundary conditions (determine integration constants)
Following table gives boundary conditions for different types of support.
Types of support and Boundary Conditions Clamped or Built in support or Fixed end : ( Point A)
Deflection, y 0 Slope, T 0 Moment , M z 0
i.e.A finite value
Free end: (Point B)
Deflection, y z 0
i.e.A finite value
Slope, T z 0 i.e.A finite value Moment , M
0
Roller (Point B) or Pinned Support (Point A) or Hinged or Simply supported.
Deflection, y 0 Slope, T z 0 i.e.A finite value Moment , M
0
End restrained against rotation but free to deflection
Deflection, y z 0 i.e.A finite value Slope, T
0
Shear force, V
0
Flexible support
Deflection, y z 0 i.e.A finite value Slope, T z 0 i .e. A finite value
Page 199 of 429
Figure
Chapter5
Deflection of Beam
Moment , M kr Shear force, V
S K Mondal’s
dy dx k. y
Using double integration method we will find the deflection and slope of the following loaded beams one by one. (i)
A Cantilever beam with point load at the free end.
(ii)
A Cantilever beam with UDL (uniformly distributed load)
(iii)
A Cantilever beam with an applied moment at free end.
(iv)
A simply supported beam with a point load at its midpoint.
(v)
A simply supported beam with a point load NOT at its midpoint.
(vi)
A simply supported beam with UDL (Uniformly distributed load)
(vii) A simply supported beam with triangular distributed load (GVL) gradually varied load. (viii) A simply supported beam with a moment at mid span. (ix)
A simply supported beam with a continuously distributed load the intensity of which at any point ‘x’ along the beam is
wx
§Sx · w sin ¨ ¸ © L ¹
(i) A Cantilever beam with point load at the free end. We will solve this problem by double integration method. For that at first we have to calculate (Mx). Consider any section XX at a distance ‘x’ from free end which is left end as shown in figure.
? Mx =  P.x We know that differential equation of elastic line
EI
d2 y dx 2
Mx
P.x
Integrating both side we get
Page 200 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s
2
dy P Â³ x dx dx 2 dy x2 or EI P. A dx 2
Â³ EI
.............(i)
Again integrating both side we get Â§ x2 Â· Â³ Â¨Â© P 2 A Â¸Â¹ dx Px 3 or EIy = ..............(ii) Ax +B 6 Where A and B is integration constants.
EIÂ³ dy =
Now apply boundary condition at fixed end which is at a distance x = L from free end and we also know that at fixed end at x = L,
y=0
at x = L,
dy dx
0
from equation (ii) EIL = 
PL3 + AL +B 6
from equation (i) EI.(0) = 
PL2 +A 2
Solving (iii) & (iv) we get A =
Therefore,
y=
..........(iii) â€¦..(iv)
PL2 PL3 and B = 2 3
Px 3 PL2 x PL3 6EI 2EI 3EI
The slope as well as the deflection would be maximum at free end hence putting x = 0 we get ymax = 
PL3 (Negative sign indicates the deflection is downward) 3EI
(Slope)max = T
max
=
PL2 2EI
Remember for a cantilever beam with a point load at free end.
3
Downward deflection at free end,
And slope at free end,
T
G
PL 3EI
PL2 2EI Page 201 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s
(ii) A Cantilever beam with UDL (uniformly distributed load)
We will now solve this problem by double integration method, for that at first we have to calculate (Mx). Consider any section XX at a distance â€˜xâ€™ from free end which is left end as shown in figure.
? Mx
w.x .
x 2
wx 2 2
We know that differential equation of elastic line
EI
d2 y dx 2
wx 2 2
Integrating both sides we get
d2 y wx 2 dx dx 2 Âł 2 dy wx 3 A ......(i) or EI dx 6
Âł EI
Again integrating both side we get Â§ wx 3 Âˇ Âł Â¨ÂŠ 6 A Â¸Âš dx wx 4 or EIy = Ax B.......(ii) 24 > where A and B are integration constants@ EIÂł dy
Now apply boundary condition at fixed end which is at a distance x = L from free end and we also know that at fixed end. at x = L,
y=0
at x = L,
dy =0 dx
from equation (i) we get
EI u (0) =
from equation (ii) we get
wL3 +wL3 + A or A = 6 6
EI.y = 
or
B=
wL4 + A.L + B 24
wL4 8
The slope as well as the deflection would be maximum at the free end hence putting x = 0, we get Page 202 of 429
Chapter5
Deflection of Beam 4
y max
wL 8EI
slope max
S K Mondal’s
>Negative sign indicates the deflection is downward@ Tmax
wL3 6EI
Remember: For a cantilever beam with UDL over its whole length,
4
Maximum deflection at free end
Maximum slope,
T
wL 8EI
G
wL3 6EI
(iii) A Cantilever beam of length ‘L’ with an applied moment ‘M’ at free end.
Consider a section XX at a distance ‘x’ from free end, the bending moment at section XX is (Mx) = M We know that differential equation of elastic line
or EI
d2 y dx 2
M
Integrating both side we get d2 y ³ M dx dx 2 dy or EI Mx + A ...(i) dx or EI³
Page 203 of 429
Chapter5 Deflection of Beam Again integrating both side we get EIÂł dy =
S K Mondalâ€™s
Âł M x +A dx
Mx 2 Ax + B ...(ii) 2 Where A and B are integration constants.
or EI y
applying boundary conditions in equation (i) &(ii) at x = L,
dy dx
0 gives A = ML
ML2 2 2 Mx MLx ML2 Therefore deflection equation is y = 2EI EI 2EI Which is the equation of elastic curve.
at x = L, y = 0 gives B =
ML2 ML2 2
2
ML G = 2EI ML T
EI
? Maximum deflection at free end
? Maximum slope at free end
(It is downward)
Let us take a funny example: A cantilever beam AB of length â€˜Lâ€™ and uniform flexural rigidity EI has a bracket BA (attached to its free end. A vertical downward force P is applied to free end C of the bracket. Find the ratio a/L required in order that the deflection of point A is zero.
[ISRO â€“ 2008]
We may consider this force â€˜Pâ€™ and a moment (P.a) act on free end A of the cantilever beam.
Page 204 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s 3
Due to point load â€˜Pâ€™ at free end â€˜Aâ€™ downward deflection G
Due to moment M = P.a at free end â€˜Aâ€™ upward deflection G
PL 3EI ML2 2EI
(P.a)L2 2EI
For zero deflection of free end A
PL3 (P.a)L2 = 3EI 2EI a 2 or L 3
(iv) A simply supported beam with a point load P at its midpoint. A simply supported beam AB carries a concentrated load P at its midpoint as shown in the figure.
We want to locate the point of maximum deflection on the elastic curve and find its value.
In the region 0 < x < L/2 Bending moment at any point x (According to the shown coordinate system)
Â§PÂˇ Â¸ .x ÂŠ2Âš
Mx = Â¨
and In the region L/2 < x < L Mx =
P x L / 2
2
We know that differential equation of elastic line
EI
d2 y dx 2
P .x 2
In the region 0 < x < L/2
Integrating both side we get d2 y
or EI
Âł dx
or EI
dy dx
2
P
Âł 2 x dx P x2 . A (i) 2 2
Again integrating both side we get
Page 205 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s
Â§P Âˇ EI Âł dy = Âł Â¨ x 2 A Â¸ dx ÂŠ4 Âš 3 Px or EI y = Ax + B (ii) 12 > Where A and B are integrating constants@ Now applying boundary conditions to equation (i) and (ii) we get
at x = 0, at x = L/2, A=
y=0 dy dx
0
PL2 and B = 0 16
? Equation of elastic line, y =
Px 3 PL12 x 12 16
3
Maximum deflection at mid span (x = L/2)
and maximum slope at each end
PL G = 48EI
T
PL2 16EI
(v) A simply supported beam with a point load â€˜Pâ€™ NOT at its midpoint. A simply supported beam AB carries a concentrated load P as shown in the figure.
We have to locate the point of maximum deflection on the elastic curve and find the value of this deflection. Taking coordinate axes x and y as shown below
Page 206 of 429
Chapter5
Deflection of Beam
For the bending moment we have In the region 0 d x d a,
Mx
And, In the region a d x d L,
Mx
Â§ P.a Âˇ Â¨ L Â¸ .x ÂŠ Âš
P.a L  x
L
So we obtain two differential equation for the elastic curve. d2 y P.a .x L dx 2 d2 y P.a and EI 2 . L  x
dx L
for 0 d x d a
EI
for a d x d L
Successive integration of these equations gives
dy P.a x 2 . + A1 ......(i) dx L 2 dy P.a 2 EI P.a x x A2 ......(ii) dx L P.a x 3 EI y . +A1x+B1 ......(iii) L 6 x 2 P.a x 3 EI y P.a . A 2 x + B2 .....(iv) 2 L 6
for o d x d a
EI
for a d x d L for 0 d x d a for a d x d L
Where A1, A2, B1, B2 are constants of Integration. Now we have to use Boundary conditions for finding constants: BCS
(a) at x = 0, y = 0 (b) at x = L, y = 0
Â§ dy Âˇ Â¸ = Same for equation (i) & (ii) ÂŠ dx Âš
(c) at x = a, Â¨
(d) at x = a, y = same from equation (iii) & (iv) We get
A1 and B1
Pb 2 L b2 ; 6L
A2
P.a 2L2 a2 6L
0;
B2
Pa3 / 6EI
Page 207 of 429
Therefore we get two equations of elastic curve
S K Mondalâ€™s
Chapter5
Deflection of Beam
Pbx 2 L b2 x 2 ..... (v) 6L Âş Pb ÂŞÂ§ L Âˇ 3 EI y = x  a L2 b2 x  x3 Âť . ...(vi) Â¨ Â¸ ÂŤ 6L ÂŹÂŠ b Âš Âź EI y = 
S K Mondalâ€™s
for 0 d x d a
for a d x d L
For a > b, the maximum deflection will occur in the left portion of the span, to which equation (v) applies. Setting the derivative of this expression equal to zero gives
x=
a(a+2b) 3
L2 b2 3
(Lb)(L+b) 3
at that point a horizontal tangent and hence the point of maximum deflection substituting this value of x into equation (v), we find,
y max
P.b(L2 b2 )3/2 9 3. EIL
Case â€“I: if a = b = L/2 then Maximum deflection will be at x =
L2 L/2
3
2
L/2
i.e. at mid point
and y max
G
^
P. L/2 u L2 L/2
2
3/2
`
9 3 EIL
PL3 48EI
(vi) A simply supported beam with UDL (Uniformly distributed load) A simply supported beam AB carries a uniformly distributed load (UDL) of intensity w/unit length over its whole span L as shown in figure. We want to develop the equation of the elastic curve and find the maximum deflection G at the middle of the span.
Taking coordinate axes x and y as shown, we have for the bending moment at any point x
Mx
wL x2 .x  w. 2 2
Then the differential equation of deflection becomes
EI
d2 y dx 2
Mx
wL x2 .x  w. 2 2
Integrating both sides we get
EI
dy dx
wL x 2 w x 3 . . A 2 2 2 3
Page 208 of 429 .....(i)
Chapter5
Deflection of Beam
S K Mondalâ€™s
Again Integrating both side we get
EI y
wL x 3 w x 4 . . Ax + B 2 6 2 12
.....(ii)
Where A and B are integration constants. To evaluate these constants we have to use boundary conditions. at x = 0, y = 0 at x = L/2,
dy dx
gives
0
gives
B=0
A
wL3 24
Therefore the equation of the elastic curve
y
wL 3 w wL3 wx ÂŞL3 2L.x 2 x 3 ÂşÂź .x .x 4 .x = 12EI 24EI 12EI 24EI ÂŹ
The maximum deflection at the midspan, we have to put x = L/2 in the equation and obtain
4
Maximum deflection at midspan,
And Maximum slope T A
G
5wL 384EI
(It is downward)
TB at the left end A and at the right end b is same putting x = 0 or x = L
Therefore we get Maximum slope
T
wL3 24EI
(vii) A simply supported beam with triangular distributed load (GVL) gradually varied load. A simply supported beam carries a triangular distributed load (GVL) as shown in figure below. We have to find equation of elastic curve and find maximum deflection G .
In this (GVL) condition, we get
Page 209 of 429
Chapter5
Deflection of Beam 4
EI
dy dx 4
load
w .x L
S K Mondalâ€™s
.....(i)
Separating variables and integrating we get
EI
d3 y dx 3
Vx
wx 2 +A 2L
.....(ii)
Again integrating thrice we get
EI
d2 y dx 2
Mx
EI
dy dx
EI y
wx 3 + Ax +B 6L
.....(iii)
wx 4 Ax 2 + +Bx + C 24L 2
.....(iv)
wx 5 Ax 3 Bx 2 + + + Cx +D 120L 6 2
.....(v)
Where A, B, C and D are integration constant. Boundary conditions
A=
at x = 0,
Mx = 0,
at x = L,
Mx = 0, y = 0 gives
y=0
wL 7wL3 , B = 0, C = , D=0 360 6
Therefore y = 
wx 7L4 10L2 x 2 3x 4 360EIL
^
To find maximum deflection G , we have
`
(negative sign indicates downward deflection)
dy =0 dx
And it gives x = 0.519 L and maximum deflection G = 0.00652
wL4 EI
(viii) A simply supported beam with a moment at midspan A simply supported beam AB is acted upon by a couple M applied at an intermediate point distance â€˜aâ€™ from the equation of elastic curve and deflection at point where the moment acted.
Considering equilibrium we get R A
M and RB L
M L
Taking coordinate axes x and y as shown, we have for bending moment In the region 0 d x d a,
Mx
M .x L
In the region a d x d L,
Mx
M xM L Page 210 of 429
So we obtain the difference equation for the elastic curve
Chapter5
Deflection of Beam 2
dy dx 2 d2 y and EI 2 dx EI
M .x L M .x M L
S K Mondalâ€™s
for 0 d x d a for a d x d L
Successive integration of these equation gives
M x2 . A1 L 2 M x2  Mx+ A 2 L 2 M x3 A1x + B1 and EI y = . L V M x 3 Mx 2 A 2 x + B2 EI y = L V 2 dy dx dy EI dx EI
....(i)
for 0 d x d a
.....(ii)
for a d x d L
......(iii)
for 0 d x d a
......(iv)
for a d x d L
Where A1, A2, B1 and B2 are integration constants. To finding these constants boundary conditions (a) at x = 0, y = 0 (b) at x = L, y = 0
Â§ dy Âˇ Â¸ = same form equation (i) & (ii) ÂŠ dx Âš
(c) at x = a, Â¨
(d) at x = a, y = same form equation (iii) & (iv)
ML Ma2 + , A2 3 2L Ma2 B2 2
A1
M.a +
B1
0,
ML Ma2 3 2L
With this value we get the equation of elastic curve
Mx 6aL  3a2 x 2 2L2 6L ? deflection of x = a, Ma y= 3aL  2a2 L2 3EIL y=
^
^
`
for 0 d x d a
`
(ix) A simply supported beam with a continuously distributed load the Â§SxÂˇ intensity of which at any point â€˜xâ€™ along the beam is wx w sin Â¨ Â¸ ÂŠ L Âš
At first we have to find out the bending moment at any point â€˜xâ€™ according to the shown coordinate system. We know that
Page 211 of 429
Chapter5 d Vx
Deflection of Beam
S K Mondalâ€™s
Â§Sx Â· w sin Â¨ Â¸ Â© L Â¹
dx
Integrating both sides we get
Â³ dV
x
or Vx
Â§Sx Â· Â³ w sin Â¨ Â¸ dx +A Â© L Â¹ wL Â§Sx Â· .cos Â¨ Â¸A S Â© L Â¹
and we also know that
d Mx
Â§Sx Â· cos Â¨ Â¸A Â© L Â¹
wL
Vx
dx
S
Again integrating both sides we get
Â wL Â½ Â§Sx Â· A Â¾ dx cos Â¨ Â® Â¸ Â© L Â¹ Â¯S Â¿
Â³ d M Â³ x
wL2
or Mx
S
Â§Sx Â· sin Â¨ Â¸ Ax +B Â© L Â¹
2
Where A and B are integration constants, to find out the values of A and B. We have to use boundary conditions
and
at x = 0,
Mx = 0
at x = L,
Mx = 0
From these we get A = B = 0. Therefore Mx
wL2
S
2
Â§Sx Â· sin Â¨ Â¸ Â© L Â¹
So the differential equation of elastic curve
EI
d2 y dx 2
wL2
Mx
S
2
Â§SxÂ· sin Â¨ Â¸ Â© L Â¹
Successive integration gives
EI
dy dx
EI y
wL3
Â§Sx Â· cos Â¨ Â¸C S Â© L Â¹ wL4 Â§Sx Â· 4 sin Â¨ Â¸ Cx D S Â© L Â¹
3
.......(i) .....(ii)
Where C and D are integration constants, to find out C and D we have to use boundary conditions at x = 0,
y=0
at x = L,
y=0
and that give C = D = 0 Therefore slope equation
dy wL3 Â§Sx Â· 3 cos Â¨ Â¸ dx S Â© L Â¹ wL4 Â§Sx Â· 4 sin Â¨ S EI Â© L Â¸Â¹
EI
and Equation of elastic curve y
(ive sign indicates deflection is downward)
Â§Sx Â· Â¸ is maximum Â© L Â¹ Page 212 of 429
Deflection will be maximum if sin Â¨
Chapter5 §Sx · sin ¨ ¸=1 © L ¹
Deflection of Beam or
S K Mondal’s
x = L/2
and Maximum downward deflection G =
WL4 (downward). S 4EI
5.5 Macaulay's Method (Use of singularity function)
x
When the beam is subjected to point loads (but several loads) this is very convenient method for determining the deflection of the beam.
x
In this method we will write single moment equation in such a way that it becomes continuous for entire length of the beam in spite of the discontinuity of loading.
x
After integrating this equation we will find the integration constants which are valid for entire length of the beam. This method is known as method of singularity constant.
Procedure to solve the problem by Macaulay’s method Step – I: Calculate all reactions and moments Step – II: Write down the moment equation which is valid for all values of x. This must contain brackets.
Step – III: Integrate the moment equation by a typical manner. Integration of (xa) will be
xa
2
2
3
§ x2 · xa so on. not ¨ ax ¸ and integration of (xa)2 will be 3 © 2 ¹
Step – IV: After first integration write the first integration constant (A) after first terms and after second time integration write the second integration constant (B) after A.x . Constant A and B are valid for all values of x.
Step – V: Using Boundary condition find A and B at a point x = p if any term in Macaulay’s method, (xa) is negative (ive) the term will be neglected.
(i) Let us take an example: A simply supported beam AB length 6m with a point load of 30 kN is applied at a distance 4m from left end A. Determine the equations of the elastic curve between each change of load point and the maximum deflection of the beam.
Answer: We solve this problem using Macaulay’s method, for that first writes the general momentum equation for the last portion of beam BC of the loaded beam.
EI
d2 y dx 2
Mx
10x 30 x  4 N.m
....(i) Page 213 of 429
By successive integration of this equation (using Macaulay’s integration rule
Chapter5 e.g
Deflection of Beam
x a
Âł x a dx
2
S K Mondalâ€™s
2
)
We get
EI
dy dx
5x 2 A 15 x4
and EI y =
2
N.m2
5 3 x Ax + B  5 (x  4)3 N.m3 3
..... ii
..... iii
Where A and B are two integration constants. To evaluate its value we have to use following boundary conditions. at x = 0, and
y=0
at x = 6m, y = 0
Note: When we put x = 0, x  4 is negativre (â€“ive) and this term will not be considered for x = 0 , so our equation will be EI y =
5 3 x Ax +B, and at x = 0 , y = 0 gives B = 0 3
But when we put x = 6, x4 is positive (+ive) and this term will be considered for x = 6, y = 0 so our equation will be EI y =
5 3 x + Ax + 0 â€“ 5 (x â€“ 4)3 3
This gives
5 3 .6 A.6 0 5(6 4)3 3 or A =  53 EI .(0) =
So our slope and deflection equation will be
EI
dy dx
5x 2  53  15 x  4
2
5 3 3 x  53x + 0  5 x  4
3
and EI y
Now we have two equations for entire section of the beam and we have to understand how we use these equations. Here if x < 4 then x â€“ 4 is negative so this term will be deleted. That so why in the region o d x d 4m we will neglect (x â€“ 4) term and our slope and deflection equation will be
EI
dy dx
5x 2 53
5 3 x  53x 3 But in the region 4m x d 6m , (x â€“ 4) is positive so we include this term and our slope and and
EI y
deflection equation will be
EI
dy dx
EI y
5x 2  53  15 x  4
2
5 3 3 x  53x  5 x  4
3
Now we have to find out maximum deflection, but we donâ€™t know at what value of â€˜xâ€™ it will be maximum. For this assuming the value of â€˜xâ€™ will be in the region 0 d x d 4m .
Page 214 of 429
Chapter5
Deflection of Beam S K Mondalâ€™s dy Deflection (y) will be maximum for that = 0 or 5x 2  53 = 0 or x = 3.25 m as our calculated x is dx in the region 0 d x d 4m ; at x = 3.25 m deflection will be maximum or
EI ymax =
or
ymax = 
5 u 3.253 â€“ 53 u 3.25 3 115 EI
(ive sign indicates downward deflection)
But if you have any doubt that Maximum deflection may be in the range of 4 x d 6m , use EIy = 5x2 â€“ 53x â€“ 5 (x â€“ 4)3 and find out x. The value of x will be absurd that indicates the maximum deflection will not occur in the region 4 x d 6m . Deflection (y) will be maximum for that
dy =0 dx
2
or
5x 2  53  15 x  4 = 0
or
10x2 120x + 293 = 0
or
x = 3.41 m or 8.6 m
Both the value fall outside the region 4 x d 6m and in this region 4 x d 6m and in this region maximum deflection will not occur.
(ii)
Now take an example where Point load, UDL and Moment applied simultaneously in a beam:
Let us consider a simply supported beam AB (see Figure) of length 3m is subjected to a point load 10 kN, UDL = 5 kN/m and a bending moment M = 25 kNm. Find the deflection of the beam at point D if flexural rigidity (EI) = 50 KNm2.
Answer: Considering equilibrium
ÂŚM
A
0 gives
10 u 1  25  5 u 1 u 1 1 1/ 2 RB u 3 or RB
0
15.83kN
R A RB
10 5 u 1 gives R A
0.83 kN
We solve this problem using Macaulayâ€™s method, for that first writing the general momentum equation for the last portion of beam, DB of the loaded beam.
d2 y EI dx 2
Mx
0.83x
0
10 x1 25 x2
5 x2
Page 215 of 429 2
2
Chapter5
Deflection of Beam
S K Mondalâ€™s
By successive integration of this equation (using Macaulayâ€™s integration rule e.g
Âł x a dx
x a
2
2
)
We get
EI
dy dx
and EIy =
0.83 2 5 2 3 .x A 5 x 1 25 x 2 x 2
2 6
0.83 3 5 25 5 3 2 4 x Ax +B  x 1 x 2 x 2
6 3 2 24
Where A and B are integration constant we have to use following boundary conditions to find out A & B. at x = 0,
y=0
at x = 3m, y = 0 Therefore B = 0
0.83 5 5 u 33 A u 3 + 0  u 23 12.5 u 12 u 14 6 3 24 or A = 1.93 and 0 = 
EIy =  0.138x3 1.93x 1.67 x 1
3
2
12.5 x 2 0.21 x 2
4
Deflextion at point D at x = 2m EIyD
0.138 u 23 1.93 u 2 1.67 u 13
or yD
8.85 EI
8.85
8.85 m ive sign indicates deflection downward
50 u 103 0.177mm downward .
(iii) A simply supported beam with a couple M at a distance â€˜aâ€™ from left end If a couple acts we have to take the distance in the bracket and this should be raised to the power zero. i.e. M(x â€“ a)0. Power is zero because (x â€“ a)0 = 1 and unit of M(x â€“ a)0 = M but we introduced the distance which is needed for Macaulayâ€™s method.
EI
d2 y dx 2
M R A. x M xa
0
Successive integration gives EI
dy dx
EI y
M x2 1 . A  M xa
L 2 M xa
M 3 x Ax + B 6L 2
2
Where A and B are integration constants, we have to use boundary conditions to find out A & B. at x = 0, y = 0 gives B = 0 at x = L, y = 0 gives A =
M La
2L
2
ML 6
Page 216 of 429
Chapter5
Deflection of Beam
S K Mondal’s
8. Moment area method x
This method is used generally to obtain displacement and rotation at a single point on a beam.
x
The moment area method is convenient in case of beams acted upon with point loads in which case bending moment area consist of triangle and rectangles. A
Loading
B
C
L Mn
Mc
B.M.diag
MB
θ
X
Deflection
θ2
OA
A
ymax
C
θ AB
tÎ
x
θ AD
tBA
A
Angle between the tangents drawn at 2 points A&B on the elastic line, ș AB
ș AB =
1 u Area of the bending moment diagram between A&B EI i.e. slope
x
B
D
T AB
A B.M. EI
Deflection of B related to 'A' yBA = Moment of
M diagram between B&A taking about B (or w.r.t. B) EI i.e. deflection yBA
A B.M u x EI
Important Note If
A1 = Area of shear force (SF) diagram
A2 = Area of bending moment (BM) diagram, Then, Change of slope over any portion of the loaded beam =
Page 217 of 429
A1 u A2 EI
Chapter5
Deflection of Beam
S K Mondalâ€™s
Some typical bending moment diagram and their area (A) and distance of C.G from one edge
x is shown in the following table. [Note the distance
will be different from other end] Shape
BM Diagram
Area
Distance from C.G
1. Rectangle
A
bh
x
b 2
x
b 3
x
b 4
2. Triangle
3. Parabola
4. Parabola 5.Cubic Parabola 6. y = k xn 7. Sine curve
Determination of Maximum slope and deflection by Moment Area Method (i) A Cantilever beam with a point load at free end
Page 218 of 429
Chapter5
Deflection of Beam
Area of BM (Bending moment diagram)
1 u L u PL 2 Therefore
A
PL2 2
Maximum slope T
A EI
Maximum deflection G
§ PL2 · § 2 · ¨ ¸ u ¨ L¸ © 2 ¹ ©3 ¹ EI
PL2 2EI Ax EI
PL3 3EI
(at free end)
(at free end)
(ii) A cantilever beam with a point load not at free end Area of BM diagram A
1 u a u Pa 2
Pa2 2
Therefore
Maximum slope T
A EI
Maximum deflection G
§ Pa2 · § a · ¨ ¸ u ¨ L ¸ © 2 ¹ © 3¹ EI
Pa2 2EI Ax EI
( at free end)
Pa2 § a · . L(at free end) 2EI ¨© 3 ¸¹
(iii) A cantilever beam with UDL over its whole length Area of BM diagram A
§ wL2 · 1 uL u ¨ ¸ 3 © 2 ¹
wL3 6
Therefore
Maximum slope T
A EI
Maximum deflection G
§ wL3 · § 3 · ¨ ¸u ¨ L¸ © 6 ¹ ©4 ¹ EI
wL3 6EI Ax EI
(at free end)
wL4 (at free end) 8EI
(iv) A simply supported beam with point load at midspam
Page 219 of 429
S K Mondal’s
Chapter5
Deflection of Beam
S K Mondal’s
Area of shaded BM diagram
A
1 L PL u u 2 2 4
PL2 16
Therefore
A EI
Maximum slope T
Maximum deflection G
§ PL2 L · u ¸ ¨ © 16 3 ¹ EI
PL2 (at each ends) 16EI Ax EI
PL3 48EI
(at mid point)
(v) A simply supported beam with UDL over its whole length Area of BM diagram (shaded)
A
2 § L · § wL2 · u u¨ ¸ 3 ¨© 2 ¸¹ © 8 ¹
wL3 24
Therefore
Maximum slope T
A EI
Maximum deflection G
§ wL3 · § 5 L · ¨ ¸u¨ u ¸ © 24 ¹ © 8 2 ¹ EI
wL3 24EI Ax EI
5 wL4 384 EI
(at each ends)
(at mid point)
9. Method of superposition Assumptions: x
Structure should be linear
x
Slope of elastic line should be very small.
x
The deflection of the beam should be small such that the effect due to the shaft or rotation of the line of action of the load is neglected.
Principle of Superposition: •
Deformations of beams subjected to combinations of loadings may be obtained as the linear combination of the deformations from the individual loadings
•
Procedure is facilitated by tables of solutions for common types of loadings and supports.
Example:
For the beam and loading shown, determine the slope and deflection at point B. Page 220 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s
Superpose the deformations due to Loading I and Loading II as shown.
10. Conjugate beam method In the conjugate beam method, the length of the conjugate beam is the same as the length of the actual beam, the loading diagram (showing the loads acting) on the conjugate beam is simply the bendingmoment diagram of the actual beam divided by the flexural rigidity EI of the actual beam, and the corresponding support condition for the conjugate beam is given by the rules as shown below.
Corresponding support condition for the conjugate beam Page 221 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s
Conjugates of Common Types of Real Beams Conjugate beams for statically determinate
Conjugate
beams
real beams
indeterminate real beams
for
Statically
By the conjugate beam method, the slope and deflection of the actual beam can be found by using the following two rules: x
The slope of the actual beam at any cross section is equal to the shearing force at the corresponding cross section of the conjugate beam.
x
The deflection of the actual beam at any point is equal to the bending moment of the conjugate beam at the corresponding point.
Procedure for Analysis x
Construct the M / EI diagram for the given (real) beam subjected to the specified (real) loading. If a combination of loading exists, you may use Mdiagram by parts
x
Determine the conjugate beam corresponding to the given real beam
x
Apply the M / EI diagram as the load on the conjugate beam as per sign convention
x
Calculate the reactions at the supports of the conjugate beam by applying equations of equilibrium and conditions
x
Determine the shears in the conjugate beam at locations where slopes is desired in the
real beam, Vconj = Ç‰real x
Determine the bending moments in the conjugate beam at locations where deflections is desired in the real beam, Mconj = yreal
Page 222 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s
The method of double integration, method of superposition, momentarea theorems, and Castiglianoâ€™s theorem are all well established methods for finding deflections of beams, but they require that the boundary conditions of the beams be known or specified. If not, all of them become helpless. However, the conjugate beam method is able to proceed and yield a solution for the possible deflections of the beam based on the support conditions, rather than the boundary conditions, of the beams.
(i) A Cantilever beam with a point load â€˜Pâ€™ at its free end. For Real Beam: At a section a distance â€˜xâ€™ from free end consider the forces to the left. Taking moments about the section gives (obviously to the left of the section) Mx = P.x (negative sign means that the moment on the left hand side of the portion is in the anticlockwise direction and is therefore taken as negative according to the sign convention) so that the maximum bending moment occurs at the fixed end i.e. Mmax =  PL ( at x = L)
Considering equilibrium we get, MA
wL2 and Reaction R A
3
wL 2
Considering any crosssection XX which is at a distance of x from the fixed end. At this point load (Wx )
Shear force Vx
W .x L
R A area of triangle ANM
wL 1 Â§ w Âˇ wL wx 2  . Â¨ .x Â¸ .x = + 2 2 ÂŠL Âš 2 2L ? The shear force variation is parabolic. wL at x 0, Vx i.e. Maximum shear force, Vmax 2 at x L, Vx 0 Bending moment Mx = R A .x 
wx 2 2x .  MA 2L 3
Page 223 of 429
wL 2
Chapter5 Deflection of Beam 3 2 wL wx wL = .x 2 6L 3 ? The bending moment variation is cubic
at x = 0, Mx
at x
0
L, Mx
wL2 i.e.Maximum B.M. Mmax
3
wL2 . 3
Page 224 of 429
S K Mondalâ€™s
Chapter5
Deflection of Beam
S K Mondal’s
OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20Years GATE Questions Beam Deflection GATE1. A lean elastic beam of given flexural rigidity, EI, is loaded by a single force F as shown in figure. How many boundary conditions are necessary to determine the deflected centre line of the beam? (a) 5
(b) 4
(c) 3
(d) 2
[GATE1999] 2
GATE1. Ans. (d) EI
dy dx 2
M . Since it is second order differential equation so we need two boundary
conditions to solve it.
Double Integration Method GATE2.
A simply supported beam carrying a concentrated load W at midspan deflects by ǅ1 under the load. If the same beam carries the load W such that it is distributed uniformly over entire length and undergoes a deflection ǅ2 at the mid span. The ratio ǅ1: ǅ2 is: (a) 2: 1
GATE2. Ans. (d) G1
GATE3.
(b)
Wl3 48EI
and G 2
[IES1995; GATE1994]
2:1 §W· 5 ¨ ¸ l4 © l ¹ 384EI
(c) 1: 1
(d) 1: 2
5Wl3 Therefore ǅ1: ǅ2 = 5: 8 384EI
A simply supported laterally loaded beam was found to deflect more than a specified value.
[GATE2003]
Which of the following measures will reduce the deflection? (a)
Increase the area moment of inertia
(b)
Increase the span of the beam
(c)
Select a different material having lesser modulus of elasticity
(d)
Magnitude of the load to be increased
GATE3. Ans. (a) Maximum deflection (ǅ) =
Wl3 48EI
To reduce, ǅ, increase the area moment of Inertia.
Page 225 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s
Previous 20Years IES Questions Double Integration Method IES1.
Consider the following statements: [IES2003] In a cantilever subjected to a concentrated load at the free end 1. The bending stress is maximum at the free end 2. The maximum shear stress is constant along the length of the beam 3. The slope of the elastic curve is zero at the fixed end Which of these statements are correct? (a) 1, 2 and 3 (b) 2 and 3 (c) 1 and 3 (d) 1 and 2 IES1. Ans. (b) IES2.
A cantilever of length L, moment of inertia I. Young's modulus E carries a concentrated load W at the middle of its length. The slope of cantilever at the free end is: [IES2001] (a)
WL2 2 EI
(b)
WL2 4 EI
(c)
WL2 8 EI
(d)
WL2 16 EI
2
Â?LÂŹ W ÂžÂž ÂÂÂ ÂžÂ&#x; 2 ÂŽ WL2 IES2. Ans. (c) R 2EI 8EI IES3.
The two cantilevers A and B shown in the figure have the same uniform crosssection and the same material. Free end deflection of cantilever 'A' is Ç…. The value of mid span deflection of the cantilever â€˜Bâ€™ is:
1 a G 2
IES3. Ans. (c) G
y mid
IES4.
2 b G 3
c G
WL3 Â§ WL2 Âˇ 5WL3 Â¨ Â¸L 3EI ÂŠ 2EI Âš 6EI W Â§ 2Lx 2 x 3 Âˇ 5WL3 Â¨ Â¸ EI ÂŠ 2 6 Âšat x L 6EI
d 2G
G
A cantilever beam of rectangular crosssection is subjected to a load W at its free end. If the depth of the beam is doubled and the load is halved, the deflection of the free end as compared to original deflection will be: [IES1999] (a) Half (b) Oneeighth (c) Onesixteenth (d) Double
IES4. Ans. (c) Deflectionin cantilever
Wl 3 3EI
Wl 3 u12 3Eah3
If h is doubled, and W is halved, New deflection IES5.
[IES2000]
4Wl 3 Eah3 4Wl 3 2 Ea 2h
3
=
1 4Wl 3 u 16 Eah3
A simply supported beam of constant flexural rigidity and length 2L carries a concentrated load 'P' at its midspan and the deflection under the load is G . If a cantilever beam of the same flexural rigidity and length 'L' is subjected to load 'P' at its free end, then the deflection at the free end will be: [IES1998]
1 a G 2
b G
c 2G
Page 226 of 429
d 4G
Chapter5
Deflection of Beam
IES5. Ans. (c)
W 2L
G for simply supported beam
48 EI
3
S K Mondalâ€™s
WL3 6 EI
3
and deflection for Cantilever IES6.
WL 3EI
Two identical cantilevers are loaded as shown in the respective figures. If slope at the free end of the cantilever in figure E is Ç‰, the slope at free and of the cantilever in figure F will be:
2G
Figure E
Figure F [IES1997]
(a)
1 T 3
(b)
1 T 2
(c)
2 T 3
IES6. Ans. (d) When a B. M is applied at the free end of cantilever,
T
(d)
ML EI
PL / 2 L EI
When a cantilever is subjected to a single concentrated load at free end, then
IES7.
PL2 2 EI PL2 2 EI
A cantilever beam carries a load W uniformly distributed over its entire length. If the same load is placed at the free end of the same cantilever, then the ratio of maximum deflection in the first case to that in the second case will be: [IES1996] (a) 3/8 (b) 8/3 (c) 5/8 (d) 8/5
IES7. Ans. (a) IES8.
T
T
Wl 3 Wl 3 y 8 EI 3EI
3 8
The given figure shows a cantilever of span 'L' subjected to a concentrated load 'P' and a moment 'M' at the free end. Deflection at the free end is given by [IES1996]
PL2 ML2 (a) 2 EI 3EI
ML2 PL3 (b) 2 EI 3EI
ML2 PL3 (c) 3EI 2 EI
ML2 PL3 (d) 2 EI 48 EI
IES8. Ans. (b) IES9.
For a cantilever beam of length 'L', flexural rigidity EI and loaded at its free end by a concentrated load W, match List I with List II and select the correct answer. [IES1996] List I List II A. Maximum bending moment 1. Wl B. Strain energy 2. Wl2/2EI C. Maximum slope 3. Wl3/3EI D. Maximum deflection 4. W2l2/6EI Codes: A B C D A B C D (a) 1 4 3 2 (b) 1 4 2 3 (c) 4 2 1 3 (d) 4 3 1 2
IES9. Ans. (b) IES10.
Maximum deflection of a cantilever beam of length â€˜lâ€™ carrying uniformly Page 227 of 429 distributed load w per unit length will be: [IES 2008]
Chapter5
Deflection of Beam (a) wl4/ (EI)
(b) w l4/ (4 EI)
S K Mondalâ€™s
(c) w l4/ (8 EI)
(d) w l4/ (384 EI)
[Where E = modulus of elasticity of beam material and I = moment of inertia of beam crosssection] IES10. Ans. (c)
IES11.
A cantilever beam of length â€˜lâ€™ is subjected to a concentrated load P at a distance of l/3 from the free end. What is the deflection of the free end of the beam? (EI is the flexural rigidity) [IES2004]
2 Pl 3 (a) 81EI
3Pl 3 (b) 81EI
14 Pl 3 (c) 81EI
15 Pl 3 (d) 81EI
Moment Area method gives us
IES11. Ans. (d)
GA
Area x EI
1 Â§ 2Pl Âˇ Â§ 2l Âˇ Â§ l 4 Âˇ u u u l 2 Â¨ÂŠ 3 Â¸Âš Â¨ÂŠ 3 Â¸Âš Â¨ÂŠ 3 9 Â¸Âš EI Pl3 2 7 14 Pl3 u u EI 9 9 81 EI 2
Alternatively Ymax
IES12.
Â§ 2l Âˇ WÂ¨ Â¸ Wa2 Â l a Â˝ ÂŠ 3 Âš Â l 2l / 3 Â˝ ÂŽ Âž ÂŽ Âž EI ÂŻ 2 6 Âż EI ÂŻ 2 6 Âż Wl3 4 9 2
u u EI 9 18 14 Wl3 81 EI
A 2 m long beam BC carries a single concentrated load at its midspan and is simply supported at its ends by two cantilevers AB = 1 m long and CD = 2 m long as shown in the figure. The shear force at end A of the cantilever AB will be (a) Zero (b) 40 kg (c) 50 kg (d) 60 kg
[IES1997]
IES12. Ans. (c) Reaction force on B and C is same 100/2 = 50 kg. And we know that shear force is same throughout its length and equal to load at free end. IES13. Assertion (A): In a simply supported beam subjected to a concentrated load P at midspan, the elastic curve slope becomes zero under the load. [IES2003] Reason (R): The deflection of the beam is maximum at midspan. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES13. Ans. (a) IES14.
At a certain section at a distance 'x' from one of the supports of a simply supported beam, the intensity of loading, bending moment and shear force arc Wx, Mx and Vx respectively. If the intensity of loading is varying continuously along the length of the beam, then the invalid relation is: [IES2000]
a Slope Qx
Mx Vx
b Vx
dM x dx
c Wx
IES14. Ans. (a) Page 228 of 429
d 2M x dx 2
d Wx
dVx dx
Chapter5 IES15.
Deflection of Beam
S K Mondal’s
The bending moment equation, as a function of distance x measured from the left end, for a simply supported beam of span L m carrying a uniformly distributed load of intensity w N/m will be given by [IES1999]
wL w 3 Lx  Lx Nm 2 2 wL 2 w 3 c M= Lx  Lx Nm 2 2
wL w 2 x  x Nm 2 2 wL 2 wLx Nm d M= x 2 2
a M=
b M=
IES15. Ans. (b) IES16.
A simply supported beam with width 'b' and depth ’d’ carries a central load W and undergoes deflection ǅ at the centre. If the width and depth are interchanged, the deflection at the centre of the beam would attain the value [IES1997]
d a G b
2
3
d b §¨ ·¸ G ©b¹
IES16. Ans. (b) Deflection at center G
d c §¨ ·¸ G ©b¹ Wl3 48EI
In second case, deflection G c
d d §¨ ·¸ ©b¹
3/2
G
Wl3 § bd3 · 48E ¨ ¸ © 12 ¹
Wl 3 48EI c
Wl 3 § db3 · 48 E ¨ ¸ © 12 ¹
Wl 3 d2 § bd 3 · b 2 48 E ¨ ¸ © 12 ¹
d2 G b2
IES17.
A simply supported beam of rectangular section 4 cm by 6 cm carries a midspan concentrated load such that the 6 cm side lies parallel to line of action of loading; deflection under the load is ǅ. If the beam is now supported with the 4 cm side parallel to line of action of loading, the deflection under the load will be: [IES1993] (a) 0.44 ǅ (b) 0.67 ǅ (c) 1.5 ǅ (d) 2.25 ǅ IES17. Ans. (d) Use above explanation IES18.
A simply supported beam carrying a concentrated load W at midspan deflects by ǅ1 under the load. If the same beam carries the load W such that it is distributed uniformly over entire length and undergoes a deflection ǅ2 at the [IES1995; GATE1994] mid span. The ratio ǅ1: ǅ2 is: (a) 2: 1
IES18. Ans. (d) G1
(b)
Wl3 48EI
2:1
and G 2
(c) 1: 1
§W· 5 ¨ ¸ l4 © l ¹ 384EI
(d) 1: 2
5Wl3 Therefore ǅ1: ǅ2 = 5: 8 384EI
Moment Area Method IES19.
Match ListI with ListII and select the correct answer using the codes given below the Lists: [IES1997] ListI ListII 1. Moment area method A. Toughness B. Endurance strength 2. Hardness C. Resistance to abrasion 3. Energy absorbed before fracture in a tension test D. Deflection in a beam 4. Fatigue loading Code: A B C D A B C D (a) 4 3 1 2 (b) 4 3 2 1 (c) 3 4 2 1 (d) 3 4 1 2 IES19. Ans. (c) Page 229 of 429
Chapter5
Deflection of Beam
S K Mondal’s
Previous 20Years IAS Questions Slope and Deflection at a Section IAS1.
Which one of the following is represented by the area of the S.F diagram from one end upto a given location on the beam? [IAS2004] (a) B.M. at the location (b) Load at the location (c) Slope at the location (d) Deflection at the location IAS1. Ans. (a)
Double Integration Method IAS2.
Which one of the following is the correct statement? If for a beam
dM dx
[IAS2007]
0 for its whole length, the beam is a cantilever:
(a) Free from any load (b) Subjected to a concentrated load at its free end (c) Subjected to an end moment (d) Subjected to a udl over its whole span IAS2. Ans. (c) udl or point load both vary with x. But if we apply Bending Moment (M) = const. and
dM dx
0
IAS3.
In a cantilever beam, if the length is doubled while keeping the crosssection and the concentrated load acting at the free end the same, the deflection at the free end will increase by [IAS1996] (a) 2.66 times (b) 3 times (c) 6 times (d) 8 times IAS3. Ans. (d)
G
PL3 3EI
3
? G fL
G ? 2 G1
§ L2 · ¨ ¸ © L1 ¹
3
8
Conjugate Beam Method IAS4.
By conjugate beam method, the slope at any section of an actual beam is equal to: [IAS2002] (a) EI times the S.F. of the conjugate beam (b) EI times the B.M. of the conjugate beam (c) S.F. of conjugate beam (d) B.M. of the conjugate beam IAS4. Ans. (c) IAS5.
I = 375 × 106 m4; l = 0.5 m E = 200 GPa Determine the stiffness of the beam shown in the above figure (a) 12 × 1010 N/m (b) 10 × 1010 N/m (c) 4 × 1010 N/m (d) 8 × 1010 N/m
[IES2002] IAS5. Ans. (c) Stiffness means required load for unit deformation. BMD of the given beam Page 230 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s
Loading diagram of conjugate beam
The deflection at the free end of the actual beam = BM of the at fixed point of conjugate beam 3 Â?1 ÂŹÂ Â? ÂŹ Â? ÂŹ Â? ÂŹ ML ÂŹÂ 2L Â?Âž WL ÂžÂžL L ÂÂ ÂžÂž 1 q L q WL ÂÂq ÂžÂžL 2L ÂÂ 3WL y ÂžÂž q L q q
q L q Â Â Âž ÂžÂ&#x; 2 ÂŽÂ Â&#x;Âž 2 ÂŽÂ Â&#x;Âž 2 2EI ÂŽÂ Â&#x;Âž 3 ÂŽÂ 2EI EI ÂŽÂ 3 Â&#x;Âž 2EI 9 6 W 2EI 2 q 200 q10 q 375 q10
Or stiffness = 3 4 q1010 N / m 3 y 3L 3 q 0.5
Page 231 of 429
Chapter5
Deflection of Beam
S K Mondalâ€™s
Previous Conventional Questions with Answers Conventional Question GATE1999 Question:
Consider the signboard mounting shown in figure below. The wind load acting perpendicular to the plane of the figure is F = 100 N. We wish to limit the deflection, due to bending, at point A of the hollow cylindrical pole of outer diameter 150 mm to 5 mm. Find the wall thickness for the pole. [Assume E = 2.0 X 1011 N/m2]
Answer:
Given: F = 100 N; d0 = 150 mm, 0.15 my = 5 mm; E = 2.0 X 1O11 N/m2 Thickness of pole, t The system of signboard mounting can be considered as a cantilever loaded at A i.e. W = 100 N and also having anticlockwise moment of M = 100 x 1 = 100 Nm at the free end. Deflection of cantilever having concentrated load at the free end,
WL3 ML2 3EI 2EI 100 u 53 100 u 53 5 u 10 3 3 u 2.0 u 1011 u I 2 u 2.0 u 1011 u I ÂŞ 100 u 53 1 100 u 53 Âş I ÂŤ Âť 5 u 10 3 ÂŹ 3 u 2.0 u 1011 2 u 2.0 u 1011 Âź y
or
But ?
5.417 u 10 6 m4
S 4 d0 di4
64 S 5.417 u 106 0.15 4 di4
64 I
or
di
0.141m or 141 mm
?
t
d0 di 2
150 141 2
4.5 mm
Conventional Question IES2003 Question:
Find the slope and deflection at the free end of a cantilever beam of length 6m as loaded shown in figure below, using method of superposition. Evaluate their numerical value using E = 200 GPa, I = 1Ă—104 m4 and W = 1 kN.
Page 232 of 429
Chapter5 Answer:
Deflection of Beam We have to theory. 1st consider
use
S K Mondalâ€™s
superposition
PL3 (3W ) q 23 8W 3EI 3EI EI 2 2 PL (3W ).2 6W È™c 2EI 2EI EI Ä¯c
Deflection at A due to this load(Ä¯1 ) = Ä¯ c Rc .(6 2) =
8W 6W 32W
q4 EI EI EI
2nd consider:
2W q 43
128W 3EI 3EI 2 (2W ) q 4 16W RB EI 2EI Deflection at A due to this load(Ä¯ 2 ) Ä¯B
=Ä¯B RB q (6 4)=
224W 3EI
.
3rd consider : W q 63 72W 3EI EI W q 62 18W RA 2EI EI
(Ä¯3 ) Ä¯ A
Apply superpositioning form ula È™=È™ A È™ B È™ c
40 q 10 3
6W 16W 18W 40W
EI EI EI EI 200 q 10 9 q 10 4
32W 224W 72W 40W 563Ã—W =
EI EI EI 3 EI 3EI 563Ã—(10 3 ) = 8.93 m m 3 q (200 q 10 9 ) q 10 4
Ä¯ Ä¯1 Ä¯ 2 Ä¯ 3
Conventional Question IES2002 Question:
If two cantilever beams of identical dimensions but made of mild steel and grey cast iron are subjected to same point load at the free end, within elastic limit, which one will deflect more and why?
Answer:
Grey cost iron will deflect more.
We know that a cantilever beam of length 'L' end load 'P' will deflect at free end (E ) =
PL3 3EI Page 233 of 429
Chapter5
Deflection of Beam
S K Mondal’s
1 E ECast Iron 125 GPa and EMild steel 200 GPa =E r
Conventional Question IES1997 Question:
Answer:
A uniform cantilever beam (EI = constant) of length L is carrying a concentrated load P at its free end. What would be its slope at the (i) Free end and (ii) Built in end P PL2 (i) Free end, ș = (ii) Builtin end,
2EI ș0
L
Page 234 of 429
6.
Bending Stress in Beam
Theory at a Glance (for IES, GATE, PSU) 6.1 Euler Bernoulliâ€™s Equation or (Bending stress formula) or Bending Equation
V y Where
M I
E R
V = Bending Stress M = Bending Moment I
= Moment of Inertia
E = Modulus of elasticity R = Radius of curvature y = Distance of the fibre from NA (Neutral axis)
6.2 Assumptions in Simple Bending Theory All of the foregoing theory has been developed for the case of pure bending i.e. constant B.M along the length of the beam. In such case
x
The shear force at each c/s is zero.
x
Normal stress due to bending is only produced.
x
Beams are initially straight
x
The material is homogenous and isotropic i.e. it has a uniform composition and its mechanical properties are the same in all directions
x
The stressstrain relationship is linear and elastic
x
Youngâ€™s Modulus is the same in tension as in compression
x
Sections are symmetrical about the plane of bending
x
Sections which are plane before bending remain plane after bending
6.3
Page 235 of 429
Chapter6
Bending Stress in Beam
V max V t
Mc1 I
V min V c
Mc2 I
S K Mondalâ€™s
(Minimum in sense of sign)
6.4 Section Modulus (Z)
I Z= y x
Z is a function of beam c/s only
x
Z is other name of the strength of the beam
x
The strength of the beam sections depends mainly on the section modulus
V
M Z
x
The flexural formula may be written as,
x
Rectangular c/s of width is "b" & depth "h" with sides horizontal, Z =
x
Square beam with sides horizontal, Z =
x
Square c/s with diagonal horizontal, Z =
x
Circular c/s of diameter "d", Z =
a3 6 a3 6 2
Sd3 32
Page 236 of 429
bh 2 6
Chapter6
Bending Stress in Beam
S K Mondalâ€™s
A log diameter "d" is available. It is proposed to cut out a strongest beam from it. Then Z=
b( d 2 b 2 ) 6
Therefore, Zmax =
bd 3 d for b = 9 3
6.5 Flexural Rigidity (EI) Reflects both
x
Stiffness of the material (measured by E)
x
Proportions of the c/s area (measured by I )
6.6 Axial Rigidity = EA 6.7 Beam of uniform strength It is one is which the maximum bending stress is same in every section along the longitudinal axis.
M D bh 2
For it
Where b = Width of beam h = Height of beam To make Beam of uniform strength the section of the beam may be varied by
x
Keeping the width constant throughout the length and varying the depth, (Most widely used)
x
Keeping the depth constant throughout the length and varying the width
x
By varying both width and depth suitably.
6.8 Bending stress due to additional Axial thrust (P). A shaft may be subjected to a combined bending and axial thrust. This type of situation arises in various machine elements. If P = Axial thrust
Then direct stress ( V d ) = P / A (stress due to axial thrust) This direct stress ( V d ) may be tensile or compressive depending upon the load P is tensile or compressive. Page 237 of 429
Chapter6
Bending Stress in Beam
And the bending stress ( V b ) =
S K Mondalâ€™s
My is varying linearly from zero at centre and extremum (minimum I
or maximum) at top and bottom fibres.
If P is compressive then
P My A I
V
x
At top fibre
x
At mid fibre
x
At bottom fibre
(compressive)
P A
V V
(compressive)
P My â€“ (compressive) A I
6.9 Load acting eccentrically to one axis x
V max
P P u e y A I
x
V min
P P u e y A I
where â€˜eâ€™ is the eccentricity at which â€˜Pâ€™ is act.
Condition for No tension in any section x
For no tension in any section, the eccentricity must not exceed
2k 2 d
[Where d = depth of the section; k = radius of gyration of c/s]
h 6
x
For rectangular section (b x h) , e d
x
For circular section of diameter â€˜dâ€™ , e d
For hollow circular section of diameter â€˜dâ€™ , e d
i.e load will be 2e
d 8
h of the middle section. 3
i.e. diameter of the kernel, 2e
D2 d 2 8D
d 4
i.e. diameter of the kernel, 2e d
Page 238 of 429
D2 d 2 . 4D
Chapter6
Bending Stress in Beam
S K Mondal’s
OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20Years GATE Questions Bending equation GATE1.
A cantilever beam has the square cross section 10mm × 10 mm. It carries a transverse load of 10 N. Considering only the bottom fibres of the beam, the correct representation of the longitudinal variation of the bending stress is:
GATE1. Ans. (a) Mx
P.x
M I
V y
or V
[GATE2005]
My I
10 u x u 0.005
0.01
4
60.(x) MPa
12 At x 0; V 0 At x 1m; V 60MPa And it is linear as V f x GATE2.
Two beams, one having square cross section and another circular crosssection, are subjected to the same amount of bending moment. If the cross sectional area as well as the material of both the beams are the same then [GATE2003] (a) Maximum bending stress developed in both the beams is the same (b) The circular beam experiences more bending stress than the square one (c) The square beam experiences more bending stress than the circular one (d) As the material is same both the beams will experience same deformation M E V My ; or V ; GATE2. Ans. (b) I U y I
V sq ? V sq
§a· M¨ ¸ ©2¹ 1 a.a3 12 V cir
6M ; a3
V cir
§d· M¨ ¸ ©2¹ S d4 64
32M S d3
4S S M 22.27M a3 a3
ª S d2 « ¬ 4
º a2 » ¼
Section Modulus GATE3.
Match the items in Columns I and II. ColumnI ColumnII 1. Cam P. Addendum Page 239 of 429 2. Beam Q. Instantaneous centre of velocity
[GATE2006]
Chapter6
Bending Stress in Beam
R. Section modulus S. Prime circle (a) P â€“ 4, Q â€“ 2, R â€“ 3, S â€“ l (c) P â€“ 3, Q â€“ 2, R â€“ 1, S â€“ 4 GATE3. Ans. (b)
S K Mondalâ€™s
3. Linkage 4. Gear (b) P â€“ 4, Q â€“ 3, R â€“ 2, S â€“ 1 (d) P â€“ 3, Q â€“ 4, R â€“ 1, S â€“ 2
Combined direct and bending stress GATE4.
For the component loaded with a force F as shown in the figure, the axial stress at the corner point P is: [GATE2008]
(a)
F (3L b) 4b3
(b)
F (3L b) 4b3
(c)
F (3L 4b) 4b3
(d)
F (3L 2b) 4b3
GATE4. Ans. (d) Total Stress = Direct stress + Stress due to Moment =
P My F F (L b ) q b
2
A I 4b 2b q ( b )3 12
Previous 20Years IES Questions Bending equation IES1.
Beam A is simply supported at its ends and carries udl of intensity w over its entire length. It is made of steel having Young's modulus E. Beam B is cantilever and carries a udl of intensity w/4 over its entire length. It is made of brass having Young's modulus E/2. The two beams are of same length and have same crosssectional area. If Ç”A and Ç”B denote the maximum bending stresses developed in beams A and B, respectively, then which one of the following is correct? [IES2005] (a) Ç”A/Ç”B (b) Ç”A/Ç”B < 1.0 (d) Ç”A/Ç”B depends on the shape of crosssection (c) Ç”A/Ç”B > 1.0 My , y and I both depends on the IES1. Ans. (d) Bending stress V
I Shape of cross sec tion so
IES2.
VA depends on the shape of cross sec tion VB
If the area of crosssection of a circular section beam is made four times, keeping the loads, length, support conditions and material of the beam unchanged, then the qualities (ListI) will change through different factors (ListII). Match the ListI with the ListII and select the correct answer using the code given below the Lists: [IES2005] ListI ListII Page 240 of 429 A. Maximum BM 1. 8
Chapter6
Bending Stress in Beam
S K Mondal’s
B. Deflection 2. 1 C. Bending Stress 3. 1/8 D. Section Modulus 4. 1/16 Codes: A B C D A (a) 3 1 2 4 (b) 2 (c) 3 4 2 1 (d) 2 IES2. Ans. (b) Diameter will be double, D = 2d. A. Maximum BM will be unaffected B. deflection ratio C. Bending stress
EI1 EI2
§d· ¨4¸ © ¹
V
4
B 4 1
C 3 3
D 1 4
1 16 M d / 2
My I
or Bending stress ratio
S d4
V2 V1
§d· ¨D¸ © ¹
3
1 8
64
D. Selection Modulus ratio
Z2 Z1
I2 y1 u y1 I1
§D· ¨d¸ © ¹
3
8
IES3.
Consider the following statements in case of beams: [IES2002] 1. Rate of change of shear force is equal to the rate of loading at a particular section 2. Rate of change of bending moment is equal to the shear force at a particular suction. 3. Maximum shear force in a beam occurs at a point where bending moment is either zero or bending moment changes sign Which of the above statements are correct? (a) 1 alone (b) 2 alone (c) 1 and 2 (d) 1, 2 and 3 IES3. Ans. (c) IES4.
Match ListI with ListII and select the correct answer using the code given below the Lists: [IES2006] ListI (State of Stress) ListII (Kind of Loading)
Codes: (a) (c) IES4. Ans. (c)
A 2 2
B 1 4
C 3 3
D 4 1
1.
Combined bending and torsion of circular shaft
2.
Torsion of circular shaft
3.
Thin cylinder pressure
4.
Tie bar subjected to tensile force
(b) (d)
Page 241 of 429
A 3 3
B 4 1
subjected
C 2 2
to
D 1 4
internal
Chapter6
Bending Stress in Beam
S K Mondal’s
Section Modulus IES5.
Two beams of equal crosssectional area are subjected to equal bending moment. If one beam has square crosssection and the other has circular section, then [IES1999] (a) Both beams will be equally strong (b) Circular section beam will be stronger (c) Square section beam will be stronger (d) The strength of the beam will depend on the nature of loading
IES5. Ans. (b) If D is diameter of circle and 'a' the side of square section, Z for circular section =
IES6.
Sd2 32
a3 ; 4 S
4
d2
4
a 2 or d
and Z for square section =
S
a
a3 6
A beam crosssection is used in two different orientations as shown in the given figure: Bending moments applied to the beam in both cases are same. The maximum bending stresses induced in cases (A) and (B) are related as: (a) V A 4V B (b) V A 2V B (c)
VB
VA
2
(d)
VA
VB [IES1997]
4 2
M
Z A .V A
Z B .V B
§b· b¨ ¸ ©2¹ 6
bd , ZA 6 b3 b3 or V A VB, 24 12
IES6. Ans. (b) Z for rectangular section is
IES7.
S
2 3
b , 24 or V A
ZB
b 2 ub 2 6
b3 12
2V B
A horizontal beam with square crosssection is simply supported with sides of the square horizontal and vertical and carries a distributed loading that produces maximum bending stress a in the beam. When the beam is placed with one of the diagonals horizontal the maximum bending stress will be: [IES1993]
(a)
1 V 2
IES7. Ans. (c) Bending stress =
(b) V
(c)
2V
(d) 2V
M Z
a3 For rectangular beam with sides horizontal and vertical, Z = 6 3 a For same section with diagonal horizontal, Z = 6 2
? Ratio of two stresses = 2 IES8.
Which one of the following combinations of angles will carry the maximum load as a column? [IES1994]
Page 242 of 429
Chapter6
Bending Stress in Beam
S K Mondalâ€™s
IES8. Ans. (a) IES9.
Assertion (A): For structures steel Ibeams preferred to other shapes. [IES1992] Reason (R): In Ibeams a large portion of their crosssection is located far from the neutral axis. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES9. Ans. (a)
Combined direct and bending stress IES10.
Assertion (A): A column subjected to eccentric load will have its stress at centroid independent of the eccentricity. [IES1994] Reason (R): Eccentric loads in columns produce torsion. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES10. Ans. (c) A is true and R is false. IES11. For the configuration of loading shown in the given figure, the stress in fibre AB is given by: [IES1995] (a) P/A (tensile)
(c)
(b)
Â§ P P.e.5 Âˇ Â¨ Â¸ (Compressive) ÂŠ A I xx Âš
IES11. Ans. (b)
Vd
P (compressive), V x A
Â§ P P.e.5 Âˇ Â¨ Â¸ (Compressive) ÂŠ A I xx Âš
(d) P/A (Compressive)
My Ix
Pky (tensile) Ix
Page 243 of 429
Chapter6 IES12.
Bending Stress in Beam
S K Mondalâ€™s
A column of square section 40 mm Ă— 40 mm, fixed to the ground carries an eccentric load P of 1600 N as shown in the figure. If the stress developed along the edge CD is â€“1.2 N/mm2, the stress along the edge AB will be: (a) (b) (c) (d)
â€“1.2 N/mm2 +1 N/mm2 +0.8 N/mm2 â€“0.8 N/mm2
[IES1999] IES12. Ans. (d) Compressive stress at CD = 1.2 N/mm2 =
or
IES13.
6e 20
0.2. Sostress at AB
P Â§ 6e Âˇ 1600 Â§ 6e Âˇ Â¨1 Â¸ Â¨1 Â¸ AÂŠ b Âš 1600 ÂŠ 20 Âš
1600 1 0.2 0.8 N/mm2 (com) 1600
A short column of symmetric crosssection made of a brittle material is subjected to an eccentric vertical load P at an eccentricity e. To avoid tensile stress in the short column, the eccentricity e should be less than or equal to: (a) h/12 (b) h/6 (c) h/3 (d) h/2
[IES2001] IES13. Ans. (b) IES14.
A short column of external diameter D and internal diameter d carries an eccentric load W. Toe greatest eccentricity which the load can have without producing tension on the crosssection of the column would be: [IES1999]
Dd (a) 8
D2 d 2 (b) 8d
D2 d 2 (c) 8D
IES14. Ans. (c) Page 244 of 429
(d)
D2 d 2 8
Chapter6
Bending Stress in Beam
S K Mondal’s
Previous 20Years IAS Questions Bending equation IAS1.
Consider the cantilever loaded as shown below:
[IAS2004]
What is the ratio of the maximum compressive to the maximum tensile stress? (a) 1.0 (b) 2.0 (c) 2.5 (d) 3.0 IAS1. Ans. (b) ǔ=
My I
V compressive, Max
ǔ tensile, max =
IAS2.
M § 2h · u ¨ ¸ at lower end of A. I © 3 ¹
M §h· u ¨ ¸ at upper end of B I ©3¹
A 0.2 mm thick tape goes over a frictionless pulley of 25 mm diameter. If E of the material is 100 GPa, then the maximum stress induced in the tape is: [IAS 1994] (a) 100 MPa (b) 200 MPa (c) 400 MPa (d) 800 MPa
IAS2. Ans. (d)
V y or V
0.2 E Here y = 2 R
0.1 mm = 0.1 x 103 m, R =
25 mm = 12.5 x 103 m 2
100 u 103 u 0.1 u 10 3 MPa = 800MPa 12.5 u 10 3
Section Modulus IAS3.
A pipe of external diameter 3 cm and internal diameter 2 cm and of length 4 m is supported at its ends. It carries a point load of 65 N at its centre. The sectional modulus of the pipe Page will245 be:of 429 [IAS2002]
Chapter6
Bending Stress in Beam (a)
65S cm3 64
IAS3. Ans. (c) Section modulus (z)
IAS4.
(d)
65S cm3 128
A Cantilever beam of rectangular crosssection is 1m deep and 0.6 m thick. If the beam were to be 0.6 m deep and 1m thick, then the beam would. [IAS1999] (a) Be weakened 0.5 times (b) Be weakened 0.6 times (c) Be strengthened 0.6 times (d) Have the same strength as the original beam because the crosssectional area remains the same
IAS4. Ans. (b)
z1
I y
and z2 ?
IAS5.
65S 65S cm3 cm3 (c) 32 96 S 4 4 3 2
65S I 64 cm3 cm3 3 y 96 2
(b)
S K Mondalâ€™s
z2 z1
0.6 u 13 0.5 I y 0.72 1.2
1.2m3
1u 0.63 0.3
0.72m3
0.6 times
A Tbeam shown in the given figure is subjected to a bending moment such that plastic hinge forms. The distance of the neutral axis from D is (all dimensions are in mm) (a) Zero (b) 109 mm (c) 125 mm (d) 170 mm
[IAS2001] IAS5. Ans. (b)
IAS6.
Assertion (A): I, T and channel sections are preferred for beams. [IAS2000] Reason(R): A beam crosssection should be such that the greatest possible amount of area is as far away from the neutral axis as possible. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true Page 246 429 IAS6. Ans. (a) Because it will increase area moment ofofinertia, i.e. strength of the beam.
Chapter6 IAS7.
Bending Stress in Beam
S K Mondalâ€™s
If the Tbeam crosssection shown in the given figure has bending stress of 30 MPa in the top fiber, then the stress in the bottom fiber would be (G is centroid) (a) Zero (b) 30 MPa (c) â€“80 MPa (d) 50 Mpa
[IAS2000] IAS7. Ans. (c)
M I
V1
V2
y1
y2
or V 2
y2 u
V1 y1
30 110 30 u 30
80 MPa
As top fibre in tension so bottom fibre will be in compression. Assertion (A): A square section is more economical in bending than the circular section of same area of crosssection. [IAS1999] Reason (R): The modulus of the square section is less than of circular section of same area of crosssection. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS8. ans. (c)
IAS8.
Bimetallic Strip IAS9.
A straight bimetallic strip of copper and steel is heated. It is free at ends. The strip, will: [IAS2002] (a) Expand and remain straight (b) Will not expand but will bend (c) Will expand and bend also (d) Twist only IAS9. Ans. (c) As expansion of copper will be more than steel.
Combined direct and bending stress IAS10.
A short vertical column having a square crosssection is subjected to an axial compressive force, centre of pressure of which passes through point R as shown in the above figure. Maximum compressive stress occurs at point (a) S (b) Q (c) R (d) P [IAS2002]
IAS10. Ans. (a) As direct and bending both the stress is compressive here. IAS11.
A strut's crosssectional area A is subjected to load P a point S (h, k) as shown in the given figure. The stress at the point Q (x, y) is: [IAS2000]
Page 247 of 429
Chapter6
Bending Stress in Beam
P Phy Pkx A Ix Iy P Phx Pky (b) A Iy Ix P Phy Pkx (c) A Iy Ix P Phx Pky (d) A Iy Ix (a)
IAS11. Ans. (b) All stress are compressive, direct stress,
P My Pky (compressive), V x (compressive) A Ix Ix Mx Phx and V y (compressive) Iy Iy
Vd
Page 248 of 429
S K Mondalâ€™s
Chapter6
Bending Stress in Beam
S K Mondalâ€™s
Previous Conventional Questions with Answers Conventional Question IES2008 Question:
A Simply supported beam AB of span length 4 m supports a uniformly distributed load of intensity q = 4 kN/m spread over the entire span and a concentrated load P = 2 kN placed at a distance of 1.5 m from left end A. The beam is constructed of a rectangular crosssection with width b = 10 cm and depth d = 20 cm. Determine the maximum tensile and compressive stresses developed in the beam to bending.
Answer:
X
2KN
4kN/M
A
B 1.5 4m
NA
RA
RB B=10cm
X
C/s
R A + RB = 2 + 4Ă—4.........(i) R A Ă—4 + 2Ă—(41.5) + (4Ă—4)Ă—2=0.......(ii) or R A = 9.25 kN, RB =18R A = 8.75 kN if 0 b x b 2.5 m
2 2(x2.5)
Mx =RB Ă—x  4x. x
=8.75x  2x 2  2x + 5 = 6.75x  2x 2 + 5
...(ii)
From (i) & (ii) we find out that bending movment at x = 2.1875 m in(i) gives maximum bending movement
dM for both the casses] dx 8.25 q 2.1875 2 q18752 9.57K 7kNm
[Just find Mmax
bh3 0.1q 0.23 6.6667 q105 m 4 12 12 Maximum distance from NA is y = 10 cm = 0.1m
Area movement of Inertia (I) =
Tmax
My (9.57 q103 )q 0.1 N 14.355MPa m2 I 6.6667 q105
Therefore maximum tensile stress in the lowest point in the beam is 14.355 MPa and maximum compressive stress in the topmost fiber of the beam is 14.355 MPa.
Conventional Question IES2007 Question:
A simply supported beam made of rolled steel joist (Isection: 450mm Ă— 200mm) has a span of 5 m and it carriers a central concentrated load W. The flanges are strengthened by two 300mm Ă— 20mm plates, one riveted to each flange over the entire length of the flanges. The second moment of area of the Page 249 of 429 joist about the principal bending axis is 35060 cm4. Calculate
Chapter6
Bending Stress in Beam (i) (ii)
S K Mondalâ€™s
The greatest central load the beam will carry if the bending stress in the 300mm/20mm plates is not to exceed 125 MPa. The minimum length of the 300 mm plates required to restrict the maximum bending stress is the flanges of the joist to 125 MPa.
Answer:
Moment of Inertia of the total section about XX (I) = moment of inertia of I â€“section + moment of inertia of the plates about XX axis. 2 Â 30 q 23 Â? 45 2 ÂŹÂ ÂŻÂ° ÂĄ Âž 35060 2 ÂĄ
30 q 2 qÂž ÂÂ Â° 101370 cm 4 ÂžÂ&#x; 2 12 2 ÂŽ Â°Âą ÂĄÂ˘
(i) Greatest central point load(W): For a simply supported beam a concentrated load at centre. WL W q 5 1.25W M= 4 4 6 8 Äą.I 125 q10 q 101370 q10
517194Nm M y 0.245 = 1.25W = 517194 or W = 413.76 kN (ii) Suppose the cover plates are absent for a distance of xmeters from each support. Then at these points the bending moment must not exceed moment of resistance of â€˜Iâ€™ section alone i.e
35060 q108
Äą.I 6 125 q10 q 178878Nm y 0.245 = Bending moment at x metres from each support
Page 250 of 429
Chapter6
Bending Stress in Beam
S K Mondalâ€™s
W q x 178878 2 41760 q x 178878 or , 2 or x 0.86464 m Hence leaving 0.86464 m from each support, for the
=
middle 5  2Ă—0.86464 = 3.27 m the cover plate should be provided. Conventional Question IES2002 Question:
Calculate:
Answer:
A beam of rectangular crosssection 50 mm wide and 100 mm deep is simply supported over a span of 1500 mm. It carries a concentrated load of 50 kN, 500 mm from the left support. (i) The maximum tensile stress in the beam and indicate where it occurs: (ii) The vertical deflection of the beam at a point 500 mm from the right support. E for the material of the beam = 2 Ă— 105 MPa.
Taking moment about L
RR q1500 50 q 500 or , RR 16.667 kN or , RL RR 50 = RL 50 16.667=33.333 kN Take a section from right R, xx at a distance x.
Bending moment (Mx ) RR .x
Therefore maximum bending moment will occur at 'c' Mmax=16.667Ă—1 KNm (i) Moment of Inertia of beam crosssection
(I )
bh3 0.050q (0.100)3 4 m = 4.1667Ă—106 m 4 12 12
Applying bending equation M Äą E I y S
or, Äą max
My I
Â? 0.001ÂŹÂ Â 2 ÂŽÂ N / m 2 200MPa 4.1667 q106
16.67q103 qÂžÂžÂžÂ&#x;
It will occure where M is maximum at point 'C' (ii) Macaulay's method for determing the deflection of the beam will be convenient as there is point load. M x EI
d2y 33.333q x 50q ( x 0.5) dx 2
Page 251 of 429
Chapter6
Bending Stress in Beam
S K Mondalâ€™s
Integrate both side we get d2 y x 2 50 33.333 q ( x 0.5)2 c1 x c2 2 dx 2 2 at x=0, y=0 gives c2 0 EI
at x=1.5, y=0 gives 0=5.556Ă—(1.5)3 8.333q13 c1 q1.5 or , c1 6.945 = EIy 5.556q x 3 8.333( x 0.5)3 6.945q1 2.43 or , y
2.43 m = 2.9167 mm[downward so ive] (2Ă—10 q10 )q (4.1667q106 ) 5
6
Conventional Question AMIE1997 Question:
Answer:
If the beam crosssection is rectangular having a width of 75 mm, determine the required depth such that maximum bending stress induced in the beam does not exceed 40 MN/m2 Given: b =75 mm =0Âˇ075 m, V max =40 MN/m2 Depth of the beam, d: Figure below shows a rectangular section of width b = 0Âˇ075 m and depth d metres. The bending is considered to take place about the horizontal neutral axis N.A. shown in the figure. The maximum bending stress occurs at the outer
d above or below the neutral axis. Any 2 My , where I fibre at a distance y from N.A. is subjected to a bending stress, V I bd3 denotes the second moment of area of the rectangular section about the N.A. i.e. . 12 d , the maximum bending stress there becomes At the outer fibres, y = 2 fibres of the rectangular section at a distance
or
Â§dÂˇ Mu Â¨ Â¸ ÂŠ2Âš V max bd3 12 bd2 M V max . 6
M bd2 6
i
(ii)
For the condition of maximum strength i.e. maximum moment M, the product bd2 must be a maximum, since V max is constant for a given material. To maximize the quantity bd2 we realise that it must be expressed in terms of one independent variable, say, b, and we may do this from the right angle triangle relationship. Page 252 of 429
Chapter6
Bending Stress in Beam 2
2
b d
D
2
or
S K Mondalâ€™s
2
D2 b 2
d
Multiplying both sides by b, we get bd2 bD2 b3 To maximize bd2 we take the first derivative of expression with respect to b and set it equal to zero, as follows:
d bd2 db
d bD2 b3 db
D2 3b2
b2 d2 3b2
d2 2b2
0
...(iii) Solving, we have, depth d 2 b This is the desired radio in order that the beam will carry a maximum moment M. It is to be noted that the expression appearing in the denominator of the right side of eqn. (i) i. e.
bd2 is the section modulus (Z) of a rectangular bar. Thus, it follows; the 6
section modulus is actually the quantity to be maximized for greatest strength of the beam. Using the relation (iii), we have d=
2 x 0Âˇ075 = 0Âˇ0106 m
Now, M = V max x Z = V max x
bd2 6
Substituting the values, we get M = 40 Ă—
V max
M Z
0.075 u 0.106
2
= 0.005618 MNm
6 0.005618 0.075 u 0.106 2 / 6
40MN / m2
Hence, the required depth d = 0Âˇ106 m = 106 mm
Page 253 of 429
7.
Shear Stress in Beam
Theory at a Glance (for IES, GATE, PSU) 1. Shear stress in bending ( W ) W =
vQ Ib
Where, V = Shear force =
dM dx c1
Q = Statical moment =
³ ydA y1
I = Moment of inertia b = Width of beam c/s.
2. Statical Moment (Q) c1
Q=
³ ydA = Shaded Area × distance of the centroid of the shaded area from the neutral axis of y1
the c/s.
3. Variation of shear stress Section
Diagram
Position
of
W max
W max Rectangular
N.A
W max =
3V 2A
W max 1.5W mean W NA Circular
N.A
W max
Page 254 of 429
4 W mean 3
Chapter7
Shear Stress in Beam
Triangular
W max 1.5W mean
h from N.A 6
Trapezoidal
Section
S K Mondalâ€™s
W NA
= 1.33 W mean
h1 2
V Âª 2 h12 Âº h Â¼ 8I Â¬
h from N.A 6
W max
Diagram
Uni form
In Flange,
ISection
( W max ) W max y
1
W max y
1
h
o 2
In Web
W max y
1
W m im y
1
o
h1 2
v Âªb(h12 h12 ) th12 Âº Â¼ 8It Â¬ vb 2 Âª h h12 Âº Â¼ 8 It Â¬
4. Variation of shear stress for some more section [Asked in different examinations] Non uniform ISection
Diagonally placed square section
Lsection
Hollow circle
Tsection
Cross
Page 255 of 429
Chapter7
Shear Stress in Beam
S K Mondal’s
5. Rectangular section W max =
x
Maximum shear stress for rectangular beam:
x
For this, A is the area of the entire cross section
x
Maximum shear occurs at the neutral axis
x
Shear is zero at the top and bottom of beam
3V 2A
6. Shear stress in beams of thin walled profile section. x
Shear stress at any point in the wall distance "s" from the free edge
A Shearing occurs here
O
B
Vx
W where Vx
Vx It
s
³ ydA o
Shear force
W = Thickness of the section I = Moment of inrertia about NA x
Shear Flow (q) s
q =
x
Wt
Vx ydA I NA ³o
Shear Force (F)
Page 256 of 429
Chapter7
Shear Stress in Beam
F= x
³
qds
Shear Centre (e) Point of application of shear stress resultant
Page 257 of 429
S K Mondal’s
Chapter7
Shear Stress in Beam
S K Mondal’s
OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20Years GATE Questions Shear Stress Variation GATE1.
The transverse shear stress acting in a beam of rectangular crosssection, subjected to a transverse shear load, is: (a) Variable with maximum at the bottom of the beam (b) Variable with maximum at the top of the beam (c) Uniform (d) Variable with maximum on the neutral axis [IES1995, GATE2008]
GATE1. Ans (d) GATE2.
W max
3 W mean 2
The ratio of average shear stress to the maximum shear stress in a beam with a square crosssection is: [GATE1994, 1998]
(a) 1
(b)
2 3
(c)
3 2
(d) 2
GATE2. Ans. (b)
W max
3 W mean 2
Previous 20Years IES Questions Shear Stress Variation IES1.
At a section of a beam, shear force is F with zero BM. The crosssection is square with side a. Point A lies on neutral axis and point B is mid way between neutral axis and top edge, i.e. at distance a/4 above the neutral axis. If U A and U B denote shear stresses at points A and B, then what is the value of U A / U B? [IES2005] (a) 0 (b) ¾ (c) 4/3 (d) None of above · a § a2 3 V 2 V u ¨ y2 ¸ .a 2© 4 WA VAy 3 V 2 4 ¹ 2 2 a3 a 4y or IES1. Ans. (c) W 2 W B 3 V °§ 2 Ib 2 a3 3 a4 ½ § a · ·° ua . 3 . ®¨ a 4 ¨ ¸ ¸ ¾ 12 ¨ 2 a °¯© © 4 ¹ ¸¹ °¿ Page 258 of 429